Pohjatunti

4

$c\left(HI\right)=\frac{1{,}00mol}{5{,}00l}=0{,}20\ \frac{mol}{dm^3}$

$\begin{array}{l|l} &HI&H_2&I_2\\ \hline c_{alku}&0{,}20&&\\ muutos&-2x&+x&+x\\ c_{tasap}&&& \end{array}$
$K_c=0{,}015=\frac{\left[H_2\right]\left[I_2\right]}{\left[HI\right]^2}=\frac{x^2}{\left(0{,}20-2x\right)^2}$
$x=0{,}019675$
$\left[HI\right]=0{,}20\ \frac{mol}{l}-2\cdot0{,}019675\ \frac{mol}{l}=0{,}16065\ \frac{mol}{l}$
$n\left(HI\right)=5l\cdot0{,}16065\ \frac{mol}{l}=0{,}80\ mol$

2

$c\left(CO_2\right)=c\left(H_2\right)=\frac{2{,}00\ mol}{3{,}00l}=0{,}666....\ \frac{mol}{l}$

$\begin{array}{l|l} &CO_2&H_2&CO&H_2O\\ \hline c_{alku}&0{,}6666...&0{,}6666...&0&0\\ muutos&-x&-x&+x&+x\\ c_{tasap}&0{,}6666...-x&0{,}6666...-x&x&x \end{array}$

$K_c=4{,}39=\frac{\left[CO\right]\left[H_2O\right]}{\left[CO_2\right]\left[H_2\right]}=\frac{x^2}{\left(0{,}6666...-x\right)^2}$

$x=0{,}45128...\ \left(tai\ x=1{,}27537\right)$

sivuutetaan toinen vaihtoehto, alkukonsentraatio ei voi mennä negatiiviseksi

$\left[CO_2\right]=\left[H_2\right]=0{,}21538...\approx0{,}22\ \frac{mol}{l}$

$\left[CO\right]=\left[H_2O\right]=0{,}45128...\approx0{,}45\ \frac{mol}{l}$

3

alussa

$c\left(HCl\right)=0{,}20\ \frac{mol}{l}$

$\left[H_3O^+\right]=0{,}20\ \frac{mol}{l}$

$n\left(HCl\right)=cV=0{,}20\ \frac{mol}{l}\cdot0{,}00025l=0{,}00005mol$

alussa pH

$-\lg\left(0{,}20\right)=0{,}6989$

lopussa

kun 250ml pullo täytetään vedellä, tilavuus kasvaa 25ml -> 250ml

$c\left(HCl\right)=\frac{n}{V}=\frac{0{,}00005mol}{0{,}0025l}=0{,}02\ \frac{mol}{l}$

lopussa pH

$-\lg\left(0{,}020\right)=1{,}6989$

$\Delta pH=1$

b)
$NH_3\left(aq\right)+H_2O\left(l\right)\xrightleftharpoons[]{}NH_4^+\left(aq\right)+OH^-\left(aq\right)$

$\left[NH_3\right]=0{,}50\ \frac{mol}{l}$

$K_b=\frac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}=1{,}8\cdot10^{-5}\ \frac{mol}{dm^3}$

$\Rightarrow1{,}8\cdot10^{-5}\frac{mol}{dm^3}=\frac{x^2}{0{,}5-x}$
$x=0{,}002991$
$\left[OH^-\right]=0{,}002991\ \frac{mol}{l}$
$pOH=2{,}5241$
$pH=14-pOH=11{,}475$