https://peda.net/id/b0e686aaf62 2019-10-24T09:01:16+03:00 https://peda.net/:static/532/peda.net.logo.bg.svg <div class="license">Tämän sivun lisenssi <a rel="license" href="https://peda.net/info">Peda.net-yleislisenssi</a></div> https://peda.net/id/e27de876377 2020-01-15T11:17:42+02:00 <div>Moniarvoinen protolyytti voi luovuttaa tai vastaanottaa useamman kuin yhden protonin</div> <div> </div> <div>Esim. Rikkihapon protolyysireaktio voidaan kirjoittaa:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=H_2SO_4%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Crightarrow%20HSO_4%5E-%5Cleft(aq%5Cright)%2BH_3O%5E%2B%5Cleft(aq%5Cright)" alt="H_2SO_4\left(aq\right)+H_2O\left(l\right)\rightarrow HSO_4^-\left(aq\right)+H_3O^+\left(aq\right)"/></div> <img src="https://math-demo.abitti.fi/math.svg?latex=HSO_4%5E-%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Cxrightleftharpoons%5B%5D%7B%7DSO_4%5E%7B2-%7D%5Cleft(aq%5Cright)%2BH_3O%5E%2B%5Cleft(aq%5Cright)" alt="HSO_4^-\left(aq\right)+H_2O\left(l\right)\xrightleftharpoons[]{}SO_4^{2-}\left(aq\right)+H_3O^+\left(aq\right)"/> <div>Esim. Sulfidi-ionin protolyysi:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=S%5E%7B2-%7D%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Crightarrow%20HS%5E-%5Cleft(aq%5Cright)%2BOH%5E-%5Cleft(aq%5Cright)" alt="S^{2-}\left(aq\right)+H_2O\left(l\right)\rightarrow HS^-\left(aq\right)+OH^-\left(aq\right)"/><br/> <img src="https://math-demo.abitti.fi/math.svg?latex=HS%5E-%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Cxrightleftharpoons%5B%5D%7B%7DH_2S%5Cleft(aq%5Cright)%2BOH%5E-%5Cleft(aq%5Cright)" alt="HS^-\left(aq\right)+H_2O\left(l\right)\xrightleftharpoons[]{}H_2S\left(aq\right)+OH^-\left(aq\right)"/></div> <div>Suolan happo- tai emäsluonne voidaan selvittää sen perusteella, kuinka vahvojea protolyyttejä (veteen verrattuna) suolan ionit ovat.</div> <div>Esim. Onko seuraavien suolojen vesiliuos hapan, neutraali vai emäksinen</div> <div>a) Kalsiumkloridi KCl</div> <div>Koska </div> <img src="https://math-demo.abitti.fi/math.svg?latex=K_b%5Cleft(Cl%5E-%5Cright)%3CK_b%5Cleft(H_2O%5Cright)" alt="K_b\left(Cl^-\right)&lt;K_b\left(H_2O\right)"/><span>, Kloridi-ioneilla ei ole emäsluonnetta.</span> <div>Liuos on neutraali<br/> <div> </div> <div>b) Ammoniumnitraatti <img src="https://math-demo.abitti.fi/math.svg?latex=NH_4NO_3" alt="NH_4NO_3"/></div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_b%5Cleft(NH_4%5E%2B%5Cright)%3EK_b%5Cleft(H_2O%5Cright)" alt="K_b\left(NH_4^+\right)&gt;K_b\left(H_2O\right)"/><!--filtered attribute: style="max-width: 100%; max-height: 1000px; vertical-align: middle; margin: 4px; padding: 3px 10px; cursor: pointer; border: 1px solid #e6f2f8; background: #edf9ff;"-->ja </div> <img src="https://math-demo.abitti.fi/math.svg?latex=K_b%5Cleft(NO_3%5E-%5Cright)%3CK_b%5Cleft(H_2O%5Cright)" alt="K_b\left(NO_3^-\right)&lt;K_b\left(H_2O\right)"/></div> <div>Liuos on hapan<br/> <div>c) Litiumkarbonaatti <img src="https://math-demo.abitti.fi/math.svg?latex=Li_2CO_3" alt="Li_2CO_3"/></div> <div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_b%5Cleft(CO_3%5E%7B2-%7D%5Cright)%3EK_b%5Cleft(H_2O%5Cright)" alt="K_b\left(CO_3^{2-}\right)&gt;K_b\left(H_2O\right)"/><!--filtered attribute: style="max-width: 100%; max-height: 1000px; vertical-align: middle; margin: 4px; padding: 3px 10px; cursor: pointer; border: 1px solid #e6f2f8; background: #edf9ff;"--></div> <div>Liuos on emäksinen.</div> </div> </div> 2020-01-15T11:17:42+02:00 https://peda.net/id/446807d4377 2020-01-15T10:41:58+02:00 Esim. Lasketaan pH puskuriliuokselle, jossa <img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BCH_3COOH%5Cright%5D%3D0%7B%2C%7D10%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="\left[CH_3COOH\right]=0{,}10\ \frac{mol}{l}"/>ja <img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BCH_3COO%5E-%5Cright%5D%3D0%7B%2C%7D10%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="\left[CH_3COO^-\right]=0{,}10\ \frac{mol}{l}"/><br/> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cbegin%7Barray%7D%7Bl%7Cl%7D%0A%26CH_3COOH%5Cleft(aq%5Cright)%26H_2O%5Cleft(l%5Cright)%26%5Cxrightleftharpoons%5B%5D%7B%7D%26CH_3COO%5E-%5Cleft(aq%5Cright)%26H_3O%5E%2B%5Cleft(aq%5Cright)%5C%5C%0A%5Chline%0Ac_%7Balku%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bl%7D%5Cright)%260%7B%2C%7D10%26%26%260%7B%2C%7D10%260%5C%5C%0Ac_%7Bmuutos%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bl%7D%5Cright)%26-x%26%26%26%2Bx%26%2Bx%5C%5C%0Ac_%7Btasap.%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bl%7D%5Cright)%260%7B%2C%7D10-x%26%26%260%7B%2C%7D10%2Bx%26x%0A%5Cend%7Barray%7D" alt="\begin{array}{l|l}&#10;&amp;CH_3COOH\left(aq\right)&amp;H_2O\left(l\right)&amp;\xrightleftharpoons[]{}&amp;CH_3COO^-\left(aq\right)&amp;H_3O^+\left(aq\right)\\&#10;\hline&#10;c_{alku}\left(\frac{mol}{l}\right)&amp;0{,}10&amp;&amp;&amp;0{,}10&amp;0\\&#10;c_{muutos}\left(\frac{mol}{l}\right)&amp;-x&amp;&amp;&amp;+x&amp;+x\\&#10;c_{tasap.}\left(\frac{mol}{l}\right)&amp;0{,}10-x&amp;&amp;&amp;0{,}10+x&amp;x&#10;\end{array}"/></div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_a%3D%5Cfrac%7B%5Cleft%5BCH_3COO%5E-%5Cright%5D%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%7B%5Cleft%5BCH_3COOH%5Cright%5D%7D%3D%5Cfrac%7B%5Cleft(0%7B%2C%7D10%2Bx%5Cright)%5Ccdot%20x%7D%7B0%7B%2C%7D10-x%7D%3D1%7B%2C%7D8%5Ccdot10%5E%7B-5%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bl%7D%5Cright)" alt="K_a=\frac{\left[CH_3COO^-\right]\left[H_3O^+\right]}{\left[CH_3COOH\right]}=\frac{\left(0{,}10+x\right)\cdot x}{0{,}10-x}=1{,}8\cdot10^{-5}\left(\frac{mol}{l}\right)"/></div> <img src="https://math-demo.abitti.fi/math.svg?latex=%5Crightarrow%5C%20x%5E2%2B0%7B%2C%7D100018x-1%7B%2C%7D8%5Ccdot10%5E%7B-6%7D%3D0" alt="\rightarrow\ x^2+0{,}100018x-1{,}8\cdot10^{-6}=0"/> <div>josta </div> <img src="https://math-demo.abitti.fi/math.svg?latex=x%3D1%7B%2C%7D799...%5Ccdot10%5E%7B-5%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bl%7D%5Cright)%3D%5Cleft%5BH_3O%5E%2B%5Cright%5D" alt="x=1{,}799...\cdot10^{-5}\left(\frac{mol}{l}\right)=\left[H_3O^+\right]"/><br/> <img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D-%5Clog1%7B%2C%7D799...%5Ccdot10%5E%7B-5%7D%5Capprox4%7B%2C%7D74" alt="pH=-\log1{,}799...\cdot10^{-5}\approx4{,}74"/><br/> <div>Määritetään saman purskuriliuoksen pH Henderson-Hasselbachin yhtälön avulla</div> <div>Kirjoitetaam hapovakion lauseke</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_a%3D%5Cfrac%7B%5Cleft%5BCH_3COO%5E-%5Cright%5D%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%7B%5Cleft%5BCH_3COOH%5Cright%5D%7D" alt="K_a=\frac{\left[CH_3COO^-\right]\left[H_3O^+\right]}{\left[CH_3COOH\right]}"/></div> <div>Ratkaistaan edellinen esimerkki soveltamalla Henderson-Hasselbalchin yhtälöä:</div> <div>Kirjoitetaan etikkahapon happovakion lauseke:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_a%3D%5Cfrac%7B%5Cleft%5BCH_3COO%5E-%5Cright%5D%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%7B%5Cleft%5BCH_3COOH%5Cright%5D%7D" alt="K_a=\frac{\left[CH_3COO^-\right]\left[H_3O^+\right]}{\left[CH_3COOH\right]}"/> </div> <div>Ratkaistaan lauseke <img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BH_3O%5E%2B%5Cright%5D" alt="\left[H_3O^+\right]"/>- konsentraation suhteen:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_a%5Ccdot%5Cleft%5BCH_3COOH%5Cright%5D%3D%5Cleft%5BCH_3COO%5E%7B%5E-%7D%5Cright%5D%5Cleft%5BH_3O%5E%2B%5Cright%5D" alt="K_a\cdot\left[CH_3COOH\right]=\left[CH_3COO^{^-}\right]\left[H_3O^+\right]"/></div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BH_3O%5E%2B%5Cright%5D%3DK_a%5Ccdot%5Cfrac%7B%5Cleft%5BCH_3COOH%5Cright%5D%7D%7B%5Cleft%5BCH_3COO%5E-%5Cright%5D%7D" alt="\left[H_3O^+\right]=K_a\cdot\frac{\left[CH_3COOH\right]}{\left[CH_3COO^-\right]}"/></div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BH_3O%5E%2B%5Cright%5D%3DK_a%5Ccdot%5Cfrac%7B%5Cleft%5BCH_3COOH%5Cright%5D%7D%7B%5Cleft%5BCH_3COO%5E-%5Cright%5D%7D%5C%20%5C%20%5C%20%5C%20%5C%20%5Cleft%7C%5Cright%7C%5Clog" alt="\left[H_3O^+\right]=K_a\cdot\frac{\left[CH_3COOH\right]}{\left[CH_3COO^-\right]}\ \ \ \ \ \left|\right|\log"/></div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=pH%3DpK_a-%5Clog%5Ccdot%5Cfrac%7B%5Cleft%5BCH_3COOH%5Cright%5D%7D%7B%5Cleft%5BCH_3COO%5E-%5Cright%5D%7D" alt="pH=pK_a-\log\cdot\frac{\left[CH_3COOH\right]}{\left[CH_3COO^-\right]}"/></div> <div>Esimerkin purskuriliuoksessa:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BCH_3COOH%5Cright%5D%3D0%7B%2C%7D10%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="\left[CH_3COOH\right]=0{,}10\ \frac{mol}{l}"/>ja<img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BCH_3COO%5E-%5Cright%5D%3D0%7B%2C%7D10%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="\left[CH_3COO^-\right]=0{,}10\ \frac{mol}{l}"/>ja </div> <img src="https://math-demo.abitti.fi/math.svg?latex=pK_a%3D-%5Clog1%7B%2C%7D8%5Ccdot10%5E%7B-5%7D%5Capprox4%7B%2C%7D7447" alt="pK_a=-\log1{,}8\cdot10^{-5}\approx4{,}7447"/> <div><img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D4%7B%2C%7D7447-%5Clog%5Cfrac%7B0%7B%2C%7D10%7D%7B0%7B%2C%7D10%7D%3D4%7B%2C%7D7447%5Capprox4%7B%2C%7D74" alt="pH=4{,}7447-\log\frac{0{,}10}{0{,}10}=4{,}7447\approx4{,}74"/><br/> <br/> <br/> <br/> <br/> </div> 2020-01-15T10:37:30+02:00 https://peda.net/id/d3877cbcfbb 2019-10-31T09:48:38+02:00 <div> <div>Laske titraustulosten perusteella titratuun hapn happovaki. Mikä happo voisi olla kyseessä?</div> </div> <div><br/> Ekvivalenttipisteessä V(NaOH)=25,0ml<br/> <br/> Kun puolet haposta oli neutraloitu (V(NaOH)=12,5 ml ja liuoksen pH ≈ 4,65), reaktioyhtälön</div> <img src="https://math-demo.abitti.fi/math.svg?latex=HA%5Cleft(aq%5Cright)%2BNaOH%5Cleft(aq%5Cright)%5Crightarrow%20NaA%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)" alt="HA\left(aq\right)+NaOH\left(aq\right)\rightarrow NaA\left(aq\right)+H_2O\left(l\right)"/> <div>Perusteella:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BHA%5Cright%5D%3D%5Cleft%5BA%5E-%5Cright%5D" alt="\left[HA\right]=\left[A^-\right]"/></div> <div>Siis hapon ja titrauksessa muodostuneen vastinemäksen konsentraatiot ovat yhtä suuret</div> <div>Hapon happovakion lausekkee perusteella:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=HA%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)%5Cxrightleftharpoons%5B%5D%7B%7DA%5E-%5Cleft(aq%5Cright)%2BH_3O%5E%2B%5Cleft(aq%5Cright)" alt="HA\left(aq\right)+H_2O\left(l\right)\xrightleftharpoons[]{}A^-\left(aq\right)+H_3O^+\left(aq\right)"/></div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_a%5Cleft(HA%5Cright)%3D%5Cfrac%7B%5Cleft%5BA%5E-%5Cright%5D%5Cleft%5BH_3O%5E%2B%5Cright%5D%7D%7B%5Cleft%5BHA%5Cright%5D%7D%3D%5Cleft%5BH_3O%5E%2B%5Cright%5D" alt="K_a\left(HA\right)=\frac{\left[A^-\right]\left[H_3O^+\right]}{\left[HA\right]}=\left[H_3O^+\right]"/></div> <div>(Hapon ja vastinemäksen konsentraatiot voidaan supostaa pois, koska ne ovat yhtä suuret)</div> <div> </div> <div>MAOL s.139:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=pH%3D-%5Clog%5Cleft%5BH_3O%5E%2B%5Cright%5D" alt="pH=-\log\left[H_3O^+\right]"/></div> <span>Tästä ratkaistuna:</span> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BH_3O%5E%2B%5Cright%5D%3D10%5E%7B-pH%7D" alt="\left[H_3O^+\right]=10^{-pH}"/></div> <div>Koska siis titrauksen puoless välissä pH oli noin 4,65, niin</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BH_3O%5E%2B%5Cright%5D%3D10%5E%7B-4%7B%2C%7D65%7D%3D2%7B%2C%7D238%5Ccdot10%5E%7B-5%7D%5Capprox2%7B%2C%7D2%5Ccdot10%5E%7B-5%7D%3DK_a%5Cleft(HA%5Cright)" alt="\left[H_3O^+\right]=10^{-4{,}65}=2{,}238\cdot10^{-5}\approx2{,}2\cdot10^{-5}=K_a\left(HA\right)"/></div> <div>Taulukkokirjan perusteella lähimpänä tätä arvoa on etikkahapon happovakio:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=K_a%5Cleft(CH_3COOH%5Cright)%3D1%7B%2C%7D8%5Ccdot10%5E%7B-5%7D%5Cleft(%5Cfrac%7Bmol%7D%7Bdm%5E3%7D%5Cright)" alt="K_a\left(CH_3COOH\right)=1{,}8\cdot10^{-5}\left(\frac{mol}{dm^3}\right)"/></div> <div>Titrattava happo voisi olla tämän perusteella etikkahappoa.<br/> <div> <div><br/> <div>Liuokselle voidaan määritää myös pOH-arvo:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=pOH%3D-%5Clog%5Cleft%5BOH%5E-%5Cright%5D" alt="pOH=-\log\left[OH^-\right]"/></div> <div>Tästä saaadaan edellä mainitulla tavalla:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=%5Cleft%5BOH%5E-%5Cright%5D%3D10%5E%7B-pOH%7D" alt="\left[OH^-\right]=10^{-pOH}"/></div> <span>Lisäksi:</span> <div><img src="https://math-demo.abitti.fi/math.svg?latex=pH%2BpOH%3D14%7B%2C%7D00" alt="pH+pOH=14{,}00"/></div> </div> </div> </div> 2019-10-31T09:41:18+02:00 https://peda.net/id/b7c63786f62 2019-10-24T09:01:27+03:00 <span>Esim. 100 ml:n suolahapponäyte titrattiin 0,10 M NaOH-liuoksella. Titauksessa Natriumhydroksidia kului 20,8 ml. Laske tämän suolahappoliuoksen konsentraatio.</span> <div><img src="https://math-demo.abitti.fi/math.svg?latex=HCl%5Cleft(aq%5Cright)%2BNaOH%5Cleft(aq%5Cright)%5Crightarrow%20NaCl%5Cleft(aq%5Cright)%2BH_2O%5Cleft(l%5Cright)" alt="HCl\left(aq\right)+NaOH\left(aq\right)\rightarrow NaCl\left(aq\right)+H_2O\left(l\right)"/></div> <div>Lasketaan kulunen natriumhydroksidin ainemäärä</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=e%5Cleft(NaOH%5Cright)%3D0%7B%2C%7D10%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="e\left(NaOH\right)=0{,}10\ \frac{mol}{l}"/></div> <br/> <img src="https://math-demo.abitti.fi/math.svg?latex=V%5Cleft(NaOH%5Cright)%3D20%7B%2C%7D8%5C%20ml%3D0%7B%2C%7D0208%5C%20l" alt="V\left(NaOH\right)=20{,}8\ ml=0{,}0208\ l"/><br/> <img src="https://math-demo.abitti.fi/math.svg?latex=c%3D%5Cfrac%7Bn%7D%7BV%7D%5C%20%5CLeftrightarrow%5C%20n%3Dc%5Ccdot%20V" alt="c=\frac{n}{V}\ \Leftrightarrow\ n=c\cdot V"/><br/> <img src="https://math-demo.abitti.fi/math.svg?latex=n%5Cleft(NaOH%5Cright)%3D0%7B%2C%7D10%5C%20%5Cfrac%7Bmol%7D%7Bl%7D%5Ccdot0%7B%2C%7D0208%5C%20l%3D0%7B%2C%7D00208%5C%20mol" alt="n\left(NaOH\right)=0{,}10\ \frac{mol}{l}\cdot0{,}0208\ l=0{,}00208\ mol"/><br/> <div>Kertoimen perusteella </div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=n%5Cleft(HCl%5Cright)%3Dn%5Cleft(NaOH%5Cright)%3D0%7B%2C%7D00208%5C%20mol" alt="n\left(HCl\right)=n\left(NaOH\right)=0{,}00208\ mol"/></div> <div>Koska suolahappoliuoksen tilavuus oli 100 ml=0,100 l:</div> <div><img src="https://math-demo.abitti.fi/math.svg?latex=c%5Cleft(HCl%5Cright)%3D%5Cfrac%7B0%7B%2C%7D00208%5C%20mol%7D%7B0%7B%2C%7D100%5C%20l%7D%3D0%7B%2C%7D0208%5C%20%5Cfrac%7Bmol%7D%7Bl%7D%5Capprox0%7B%2C%7D21%5C%20%5Cfrac%7Bmol%7D%7Bl%7D" alt="c\left(HCl\right)=\frac{0{,}00208\ mol}{0{,}100\ l}=0{,}0208\ \frac{mol}{l}\approx0{,}21\ \frac{mol}{l}"/></div> 2019-10-24T09:01:27+03:00