KE5S
https://peda.net/id/3c4b3e5831f
2020-01-23T11:40:19+02:00
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5.5
https://peda.net/id/14b61f4e384
2020-01-16T19:34:51+02:00
<div>
<div>5.31</div>
</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=CO_3%5E%7B2-%7D%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Cxrightleftharpoons%5B%5D%7B%7DHCO_3%5E-%5Cleft(aq%5Cright)%2BOH%5E-%5Cleft(aq%5Cright)" alt="CO_3^{2-}\left(aq\right)+H_2O\left(l\right)\xrightleftharpoons[]{}HCO_3^-\left(aq\right)+OH^-\left(aq\right)"/></div>
<br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=%5Cbegin%7Barray%7D%7Bl%7Cl%7D%0A%26CO_3%5E%7B2-%7D%26H_2O%5Cleft(l%5Cright)%26%5Cxrightleftharpoons%5B%5D%7B%7D%26HCO_3%5E-%5Cleft(aq%5Cright)%26OH%5E-%5Cleft(aq%5Cright)%5C%5C%0A%5Chline%0Ac_%7Balku%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)%260%7B%2C%7D20%26-%26%260%260%5C%5C%0Amuutos%5C%20%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)%26-x%26-%26%26%2Bx%26%2Bx%5C%5C%0Ac_%7Btasap%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)%260%7B%2C%7D20-x%26-%26%26x%26x%0A%5Cend%7Barray%7D" alt="\begin{array}{l|l} &CO_3^{2-}&H_2O\left(l\right)&\xrightleftharpoons[]{}&HCO_3^-\left(aq\right)&OH^-\left(aq\right)\\ \hline c_{alku}\left(\frac{mol}{dm^3}\right)&0{,}20&-&&0&0\\ muutos\ \left(\frac{mol}{dm^3}\right)&-x&-&&+x&+x\\ c_{tasap}\left(\frac{mol}{dm^3}\right)&0{,}20-x&-&&x&x \end{array}"/><br/>
<div>Sijoitetaan tasapainokonsentraatiot karbonaatti-ionin emäsvakion lausekkeeseen ja ratkaistaan x:</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=K_b%5Cleft(CO_3%5E%7B2-%7D%5Cright)%3D10%5E%7B-3%7B%2C%7D68%7D%3D2%7B%2C%7D1%5Ccdot10%5E%7B-4%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="K_b\left(CO_3^{2-}\right)=10^{-3{,}68}=2{,}1\cdot10^{-4}\ \frac{mol}{dm^3}"/></div>
<span>Sijoitetaan tasapainokonsentraatiot vetykarbonaatti-ionin emäsvakion lausekkeeseen ja ratkaistaan x:</span>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5CRightarrow%20K_b%3D%5Cfrac%7B%5Cleft%5BHCO_3%5E-%5Cright%5D%5Cleft%5BOH%5E-%5Cright%5D%7D%7B%5Cleft%5BCO_3%5E%7B2-%7D%5Cright%5D%7D%5CRightarrow2%7B%2C%7D1%5Ccdot10%5E%7B-4%7D%3D%5Cfrac%7Bx%5E2%7D%7B0%7B%2C%7D20-x%7D" alt="\Rightarrow K_b=\frac{\left[HCO_3^-\right]\left[OH^-\right]}{\left[CO_3^{2-}\right]}\Rightarrow2{,}1\cdot10^{-4}=\frac{x^2}{0{,}20-x}"/></div>
<div>Yhtälön ratkaisuna saadaan: </div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=x_1%3D6%7B%2C%7D377%5Ccdot10%5E%7B-3%7D" alt="x_1=6{,}377\cdot10^{-3}"/>ja <img src="https://math-demo.abitti.fi/math.svg?latex=x_2%3D-6%7B%2C%7D587%5Ccdot10%5E%7B-3%7D" alt="x_2=-6{,}587\cdot10^{-3}"/></div>
<div>Mäistä vain <img src="https://math-demo.abitti.fi/math.svg?latex=x_1" alt="x_1"/>positiivisena lukuna kelpaa, sillä hydroksidi-ionikonsentraation tulee kasvaa (ja karbonaatti-ionikonsentraation pienentyä)</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D6%7B%2C%7D377%5Ccdot10%5E%7B-3%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[OH^-\right]=6{,}377\cdot10^{-3}\ \frac{mol}{dm^3}"/></div>
<div>Lasketaan, kuink apaljon hydroksidi-ioneja muodostuu toisessa protolyysireaktiossa:</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cbegin%7Barray%7D%7Bl%7Cl%7D%0A%26HCO_3%5E-%5Cleft(aq%5Cright)%26H_2O%5Cleft(l%5Cright)%26%5Cxrightleftharpoons%5B%5D%7B%7D%26H_2SO_3%5Cleft(aq%5Cright)%26OH%5E-%5Cleft(aq%5Cright)%5C%5C%0A%5Chline%0Ac_%7Balku%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)%266%7B%2C%7D377%5Ccdot10%5E%7B-3%7D%26-%26%260%266%7B%2C%7D377%5Ccdot10%5E%7B-3%7D%5C%5C%0Amuutos%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)%26-x%26-%26%26%2Bx%26%2Bx%5C%5C%0Ac_%7Btasap%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)%266%7B%2C%7D377%5Ccdot10%5E%7B-3%7D-x%26-%26%26x%266%7B%2C%7D377%5Ccdot10%5E%7B-3%7D%2Bx%0A%5Cend%7Barray%7D" alt="\begin{array}{l|l} &HCO_3^-\left(aq\right)&H_2O\left(l\right)&\xrightleftharpoons[]{}&H_2SO_3\left(aq\right)&OH^-\left(aq\right)\\ \hline c_{alku}\left(\frac{mol}{dm^3}\right)&6{,}377\cdot10^{-3}&-&&0&6{,}377\cdot10^{-3}\\ muutos\left(\frac{mol}{dm^3}\right)&-x&-&&+x&+x\\ c_{tasap}\left(\frac{mol}{dm^3}\right)&6{,}377\cdot10^{-3}-x&-&&x&6{,}377\cdot10^{-3}+x \end{array}"/><br/>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=K_b%5Cleft(HCO_3%5E-%5Cright)%3D10%5E%7B-7%7B%2C%7D63%7D%3D2%7B%2C%7D3%5Ccdot10%5E%7B-8%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="K_b\left(HCO_3^-\right)=10^{-7{,}63}=2{,}3\cdot10^{-8}\ \frac{mol}{dm^3}"/></div>
Sijoitetaan tasapainokonsentraatiot vetykarbonaatti-ionin emäsvakion lausekkeeseen ja ratkaistaan x:</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=K_b%3D%5Cfrac%7B%5C%20%5Cleft%5BH_2CO_3%5Cright%5D%5Cleft%5BOH%5E-%5Cright%5D%7D%7B%5Cleft%5BHCO_3%5E-%5Cright%5D%7D%5CRightarrow2%7B%2C%7D3%5Ccdot10%5E%7B-8%7D%3D%5Cfrac%7Bx%5Ccdot%5Cleft(6%7B%2C%7D377%5Ccdot10%5E%7B-3%7D%2Bx%5Cright)%7D%7B6%7B%2C%7D377%5Ccdot10%5E%7B-3%7D-x%7D" alt="K_b=\frac{\ \left[H_2CO_3\right]\left[OH^-\right]}{\left[HCO_3^-\right]}\Rightarrow2{,}3\cdot10^{-8}=\frac{x\cdot\left(6{,}377\cdot10^{-3}+x\right)}{6{,}377\cdot10^{-3}-x}"/><br/>
<div>Yhtälön ratkaisuna saadaan:</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=x_1%3D2%7B%2C%7D300%5Ccdot10%5E%7B-8%7D" alt="x_1=2{,}300\cdot10^{-8}"/>ja <img src="https://math-demo.abitti.fi/math.svg?latex=x_2%3D-6%7B%2C%7D377%5Ccdot10%5E%7B-3%7D" alt="x_2=-6{,}377\cdot10^{-3}"/></div>
<div>Näistä vain <img src="https://math-demo.abitti.fi/math.svg?latex=x_1" alt="x_1"/> positiivisena lukuna kelpaa, sillä hydroksidi-ionikonsentraation tulee kasvaa (ja karbonaatti-ionikonsentraation pienentyä)</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D2%7B%2C%7D300%5Ccdot10%5E%7B-8%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[OH^-\right]=2{,}300\cdot10^{-8}\ \frac{mol}{dm^3}"/></div>
<div>Hydroksidi-ionien kokonaiskonsentraatio on </div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D6%7B%2C%7D377%5Ccdot10%5E%7B-3%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%2B2%7B%2C%7D300%5Ccdot10%5E%7B-8%7Dmol%3D6%7B%2C%7D377%5Ccdot10%5E%7B-3%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[OH^-\right]=6{,}377\cdot10^{-3}\ \frac{mol}{dm^3}+2{,}300\cdot10^{-8}mol=6{,}377\cdot10^{-3}\ \frac{mol}{dm^3}"/></div>
Huomataan, että vetykarbonaatti-ioni on huomattavasti heikompi emäs kuin karbonaatti-ioni, joten vain karbonaatti-ionin tuottama hydroksidi-ionikonsentraatio vaikuttaa liuoksen pH-arvoon.</div>
<div>pOH=2,14954</div>
<div>Liuoksen pH-arvo on:</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=pH%3DpK_w-pOH%3D13%7B%2C%7D9965-2%7B%2C%7D1954%3D11%7B%2C%7D8011%5Capprox11%7B%2C%7D80" alt="pH=pK_w-pOH=13{,}9965-2{,}1954=11{,}8011\approx11{,}80"/><br/>
<div>
<div>5.33</div>
</div>
<div>
<div>5.34</div>
<div>5.35</div>
<div>5.36</div>
<div>5.37</div>
</div>
</div>
2020-01-16T11:39:24+02:00
Kurssi 5 luku 5
https://peda.net/id/62944fb0320
2020-01-09T19:37:07+02:00
<div>5.2 Vahvan hapon tai emäksen vesiluoksen pH</div>
<div>Vahvan hapon vesiluoksen pH:n laskeminen </div>
<div>Esim. Laske suolahappolioksen pH kun liuoksen kosnetraatio on 0,5 mol/l</div>
<div> </div>
<span>Suolahapon protolyysireaktio</span>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=HCl%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Crightarrow%20Cl%5E-%5Cleft(aq%5Cright)%3DH_3O%5E%2B%5Cleft(aq%5Cright)" alt="HCl\left(aq\right)+H_2O\left(l\right)\rightarrow Cl^-\left(aq\right)=H_3O^+\left(aq\right)"/></div>
<div>Koska kyseessä on vahva happo, protolysoitumienn tapahtuu täydellisesti, jolloin</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BH_3O%5E%2B%5Cright%5D%3D%5Cleft%5BHCl%5Cright%5D%3D0%7B%2C%7D5%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="\left[H_3O^+\right]=\left[HCl\right]=0{,}5\ \frac{mol}{l}"/></div>
<img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D-%5Clog%5Cleft%5BH_3O%5E%2B%5Cright%5D%3D-%5Clog0%7B%2C%7D5%3D0%7B%2C%7D301...%5Capprox0%7B%2C%7D30" alt="pH=-\log\left[H_3O^+\right]=-\log0{,}5=0{,}301...\approx0{,}30"/><br/>
<div> </div>
<div>Vahvan emäksen vesiliuoksen pH:n laskeminen</div>
<div>Esim. Laske syntyvän liuoksen pH, kun 0,0185 grammaa kalsiumhydroksidia luotetaan veteen siten, että liuoksen lopputilavuudeksi tulee 0,50l</div>
<div> </div>
<div>Kalsiumhydroksidin liukenemisyhtälö</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=Ca%5Cleft(OH%5Cright)_2%5Cleft(s%5Cright)%5Crightarrow%20Ca%5E%7B2%2B%7D%5Cleft(aq%5Cright)%2B2OH%5E-%5Cleft(aq%5Cright)" alt="Ca\left(OH\right)_2\left(s\right)\rightarrow Ca^{2+}\left(aq\right)+2OH^-\left(aq\right)"/></div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=m%5Cleft(Ca%5Cleft(OH%5Cright)_2%5Cright)%3D0%7B%2C%7D0185g%7B%2C%7D%5C%20M%5Cleft(Ca%5Cleft(OH%5Cright)_2%5Cright)%3D74%7B%2C%7D096%5C%20%5Cfrac%7Bg%7D%7Bmol%7D" alt="m\left(Ca\left(OH\right)_2\right)=0{,}0185g{,}\ M\left(Ca\left(OH\right)_2\right)=74{,}096\ \frac{g}{mol}"/></div>
<img src="https://math-demo.abitti.fi/math.svg?latex=n%3D%5Cfrac%7Bm%7D%7BM%7D%3A" alt="n=\frac{m}{M}:"/>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=n%5Cleft(Ca%5Cleft(OH%5Cright)_2%5Cright)%3D%5Cfrac%7B0%7B%2C%7D0185g%7D%7B74%7B%2C%7D096%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%3D2%7B%2C%7D496...%5Ccdot10%5E%7B-4%7Dmol" alt="n\left(Ca\left(OH\right)_2\right)=\frac{0{,}0185g}{74{,}096\ \frac{g}{mol}}=2{,}496...\cdot10^{-4}mol"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=c%3D%5Cfrac%7Bn%7D%7BV%7D%3A" alt="c=\frac{n}{V}:"/></div>
<img src="https://math-demo.abitti.fi/math.svg?latex=c%5Cleft(Ca%5Cleft(OH%5Cright)_2%5Cright)%3D%5Cfrac%7B2%7B%2C%7D496...%5Ccdot10%5E%7B-4%7Dmol%7D%7B0%7B%2C%7D50l%7D%3D4%7B%2C%7D99...%5Ccdot10%5E%7B-4%7D%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="c\left(Ca\left(OH\right)_2\right)=\frac{2{,}496...\cdot10^{-4}mol}{0{,}50l}=4{,}99...\cdot10^{-4}\ \frac{mol}{l}"/><br/>
<div>Liukenemisyhtälön kertoimien perusteella:</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH_%2B%5Cright%5D%3D2%5Ccdot%20c%5Cleft(Ca%5Cleft(OH%5Cright)_2%5Cright)%3D2%5Ccdot4%7B%2C%7D99...%5Ccdot10%5E%7B-4%7D%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="\left[OH_+\right]=2\cdot c\left(Ca\left(OH\right)_2\right)=2\cdot4{,}99...\cdot10^{-4}\ \frac{mol}{l}"/></div>
<img src="https://math-demo.abitti.fi/math.svg?latex=%3D9%7B%2C%7D987..%5Ccdot10%5E%7B-4%7D%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="=9{,}987..\cdot10^{-4}\ \frac{mol}{l}"/><br/>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=pOH%3D-%5Clog%5Cleft%5BOH%5E-%5Cright%5D%3D-%5Clog9%7B%2C%7D987...%5Ccdot10%5E%7B-4%7D%3D3%7B%2C%7D0005..." alt="pOH=-\log\left[OH^-\right]=-\log9{,}987...\cdot10^{-4}=3{,}0005..."/></div>
<img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D14%7B%2C%7D00-3%7B%2C%7D00%3D11%7B%2C%7D00" alt="pH=14{,}00-3{,}00=11{,}00"/><br/>
<div> </div>
<div>5-1</div>
<div>a)</div>
<div>b) Endoterminen, koska kuvaaja on kasvava</div>
<div>c) </div>
<img src="https://math-demo.abitti.fi/math.svg?latex=%5Capprox1%7B%2C%7D464%5Ccdot10%5E%7B-14%7D%5C%20%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)%5E2" alt="\approx1{,}464\cdot10^{-14}\ \left(\frac{mol}{dm^3}\right)^2"/><br/>
<div>d)</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=K_w%3D%5Cleft%5BH_3O%5E%2B%5Cright%5D%5Cleft%5BOH%5E-%5Cright%5D" alt="K_w=\left[H_3O^+\right]\left[OH^-\right]"/></div>
<img src="https://math-demo.abitti.fi/math.svg?latex=K_w%3Dx%5Ccdot%20x" alt="K_w=x\cdot x"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=K_w%3Dx%5E2" alt="K_w=x^2"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=5%7B%2C%7D476%5Ccdot10%5E%7B-14%7D%3Dx%5E2" alt="5{,}476\cdot10^{-14}=x^2"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=x%3D%5Csqrt%5B%5D%7B5%7B%2C%7D476%5Ccdot10%5E%7B-14%7D%7D" alt="x=\sqrt[]{5{,}476\cdot10^{-14}}"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=x%3D2%7B%2C%7D34008%5Ccdot10%5E%7B-7%7D" alt="x=2{,}34008\cdot10^{-7}"/>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D-%5Clg2%7B%2C%7D34008%5Ccdot10%5E%7B-7%7D%3D6%7B%2C%7D6307...%5Capprox6%7B%2C%7D6" alt="pH=-\lg2{,}34008\cdot10^{-7}=6{,}6307...\approx6{,}6"/></div>
<div>e)</div>
<div>Autoprotolyysireaktio tuottaa saman määrän oksonium ja hydroksidi-ioneita.</div>
<div> </div>
<div>5-2
<div>a)</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D%5Cfrac%7BK_w%7D%7B%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%3D%5Cfrac%7B1%7B%2C%7D008%5Ccdot10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%7B1%7B%2C%7D0%5Ccdot10%5E%7B-7%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%3D1%7B%2C%7D008%5Ccdot10%5E%7B-7%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[OH^-\right]=\frac{K_w}{\left[H_3O^+\right]}=\frac{1{,}008\cdot10^{-14}\frac{mol}{dm^3}}{1{,}0\cdot10^{-7}\ \frac{mol}{dm^3}}=1{,}008\cdot10^{-7}\ \frac{mol}{dm^3}"/></div>
<div>Neutraali</div>
<span>b)</span>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D%5Cfrac%7BK_w%7D%7B%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%3D%5Cfrac%7B1%7B%2C%7D008%5Ccdot10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%7B1%7B%2C%7D9%5Ccdot10%5E%7B-11%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%3D5%7B%2C%7D3052...%5Ccdot10%5E%7B-4%7D%5Capprox5%7B%2C%7D305%5Ccdot10%5E%7B-4%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[OH^-\right]=\frac{K_w}{\left[H_3O^+\right]}=\frac{1{,}008\cdot10^{-14}\frac{mol}{dm^3}}{1{,}9\cdot10^{-11}\ \frac{mol}{dm^3}}=5{,}3052...\cdot10^{-4}\approx5{,}305\cdot10^{-4}\ \frac{mol}{dm^3}"/></div>
<div>Emäksinen</div>
<div>c)</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D%5Cfrac%7BK_w%7D%7B%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%3D%5Cfrac%7B1%7B%2C%7D008%5Ccdot10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%7B6%7B%2C%7D7%5Ccdot10%5E%7B-4%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%3D1%7B%2C%7D50447...%5Ccdot10%5E%7B-11%7D%5Capprox1%7B%2C%7D504%5Ccdot10%5E%7B-11%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[OH^-\right]=\frac{K_w}{\left[H_3O^+\right]}=\frac{1{,}008\cdot10^{-14}\frac{mol}{dm^3}}{6{,}7\cdot10^{-4}\ \frac{mol}{dm^3}}=1{,}50447...\cdot10^{-11}\approx1{,}504\cdot10^{-11}\ \frac{mol}{dm^3}"/></div>
<div>Hapan</div>
<div>d)</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D%5Cfrac%7BK_w%7D%7B%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%3D%5Cfrac%7B1%7B%2C%7D008%5Ccdot10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%7B2%7B%2C%7D3%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%7D%3D4%7B%2C%7D3826...%5Ccdot10%5E%7B-15%7D%5Capprox4%7B%2C%7D383%5Ccdot10%5E%7B-15%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[OH^-\right]=\frac{K_w}{\left[H_3O^+\right]}=\frac{1{,}008\cdot10^{-14}\frac{mol}{dm^3}}{2{,}3\ \frac{mol}{dm^3}}=4{,}3826...\cdot10^{-15}\approx4{,}383\cdot10^{-15}\ \frac{mol}{dm^3}"/></div>
<div>Hapan</div>
</div>
<div>5-3</div>
<div>5-4</div>
2020-01-08T12:48:20+02:00