5.4 Puskuriliuosten pH-arvon laskeminen
https://peda.net/id/0275c35819c
2020-10-29T10:23:23+02:00
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<div class="license">Tämän sivun lisenssi <a rel="license" href="https://peda.net/info">Peda.net-yleislisenssi</a></div>
5.25
https://peda.net/id/fca2ef1a19d
2020-10-29T12:39:13+02:00
<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-" alt="CO_3^{2-}\left(aq\right)+H_2O\left(l\right)\xrightleftharpoons[]{}HCO_3^-\left(aq\right)+OH^-"/></div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D7%7B%2C%7D30" alt="pH=7{,}30"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=pOH%3DpK_w-pH" alt="pOH=pK_w-pH"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=pK_w%3D-%5Clg%20K_w" alt="pK_w=-\lg K_w"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=K_w%3D1%7B%2C%7D008%5Ccdot10%5E%7B-14%7D" alt="K_w=1{,}008\cdot10^{-14}"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=pOH%3D6%7B%2C%7D6965" alt="pOH=6{,}6965"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D10%5E%7B-pOH%7D%3D2%7B%2C%7D011%5Ccdot10%5E%7B-7%7D" alt="\left[OH^-\right]=10^{-pOH}=2{,}011\cdot10^{-7}"/></div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=K_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%3D%5Cfrac%7B%5Cleft%5BHCO_3%5E-%5Cright%5D%5Ccdot2%7B%2C%7D011%5Ccdot10%5E%7B-7%7D%7D%7B2%7B%2C%7D0%5Ccdot10%5E%7B-7%7D%7D%3D" alt="K_b=\frac{\left[HCO_3^-\right]\left[OH^-\right]}{\left[CO_3^{2-}\right]}=\frac{\left[HCO_3^-\right]\cdot2{,}011\cdot10^{-7}}{2{,}0\cdot10^{-7}}="/></div>
2020-10-29T12:39:13+02:00
5.24
https://peda.net/id/8f64482819c
2020-10-29T11:17:25+02:00
<div>a)<br/>
<div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=HCOOH%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Cxrightleftharpoons%5B%5D%7B%7DHCOO%5E-%5Cleft(aq%5Cright)%2BH_3O%5E%2B%5Cleft(aq%5Cright)" alt="HCOOH\left(aq\right)+H_2O\left(l\right)\xrightleftharpoons[]{}HCOO^-\left(aq\right)+H_3O^+\left(aq\right)"/></div>
<div>kun lisätään happoa, tasapainoyhtälö siirtyy lähtöaineiden suuntaan</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=HCOOH%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Crightarrow%20HCOO%5E-%5Cleft(aq%5Cright)%2BH_3O%5E%2B%5Cleft(aq%5Cright)" alt="HCOOH\left(aq\right)+H_2O\left(l\right)\rightarrow HCOO^-\left(aq\right)+H_3O^+\left(aq\right)"/></div>
<div>kun lisätään emästä, tasapainoyhtälö siirtyy reaktiotuotteiden suuntaan</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=HCOO%5E-%5Cleft(aq%5Cright)%2BH_3O%5E%2B%5Cleft(aq%5Cright)%5Crightarrow%20HCOOH%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)" alt="HCOO^-\left(aq\right)+H_3O^+\left(aq\right)\rightarrow HCOOH\left(aq\right)+H_2O\left(l\right)"/></div>
</div>
b)<br/>
lasketaan alkukonsentraatiot</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=c%5Cleft(HCOOH%5Cright)%3D%5Cfrac%7B15mmol%7D%7B75ml%7D%3D0%7B%2C%7D2%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="c\left(HCOOH\right)=\frac{15mmol}{75ml}=0{,}2\ \frac{mol}{l}"/></div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=c%5Cleft(NaHCOO%5Cright)%3D%5Cfrac%7B8%7B%2C%7D5mmol%7D%7B75ml%7D%3D0%7B%2C%7D11333...%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="c\left(NaHCOO\right)=\frac{8{,}5mmol}{75ml}=0{,}11333...\ \frac{mol}{l}"/></div>
<div>tasapainotilassa oksoniumionikonsentraatio on yhtä suuri kuin <img src="https://math-demo.abitti.fi/math.svg?latex=c%5Cleft(NaHCOO%5Cright)" alt="c\left(NaHCOO\right)"/></div>
<div>taulukko</div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cbegin%7Barray%7D%7Bl%7Cl%7D%0A%26HCOOH%5Cleft(aq%5Cright)%26H_2O%5Cleft(l%5Cright)%26HCOO%5E-%5Cleft(aq%5Cright)%26H_3O%5E%2B%5Cleft(aq%5Cright)%5C%5C%0A%5Chline%0Ac_%7Balku%7D%260%7B%2C%7D2%26%260%7B%2C%7D1133...%260%5C%5C%0Amuutos%26-x%26%26%2Bx%26%2Bx%5C%5C%0Ac_%7Btasap%7D%26%26%26%26%0A%5Cend%7Barray%7D" alt="\begin{array}{l|l} &HCOOH\left(aq\right)&H_2O\left(l\right)&HCOO^-\left(aq\right)&H_3O^+\left(aq\right)\\ \hline c_{alku}&0{,}2&&0{,}1133...&0\\ muutos&-x&&+x&+x\\ c_{tasap}&&&& \end{array}"/></div>
<div><img src="https://math-demo.abitti.fi/math.svg?latex=K_a%3D%5Cfrac%7B%5Cleft%5BHCOO%5E-%5Cright%5D%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%7B%5Cleft%5BHCOOH%5Cright%5D%7D%7B%2C%7D%5C%20x%3D0%7B%2C%7D000032" alt="K_a=\frac{\left[HCOO^-\right]\left[H_3O^+\right]}{\left[HCOOH\right]}{,}\ x=0{,}000032"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BH_3O%5E%2B%5Cright%5D%3D3%7B%2C%7D2%5Ccdot10%5E%7B-5%7D%5C%20%5Cfrac%7Bmol%7D%7Bdm%5E3%7D" alt="\left[H_3O^+\right]=3{,}2\cdot10^{-5}\ \frac{mol}{dm^3}"/><br/>
<img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D-%5Clg%5Cleft%5BH_3O%5E%2B%5Cright%5D%3D4%7B%2C%7D49850...%5Capprox4%7B%2C%7D50" alt="pH=-\lg\left[H_3O^+\right]=4{,}49850...\approx4{,}50"/></div>
2020-10-29T11:17:25+02:00