Kpl.2.2

2.8
a)

$NHO_3\left(aq\right)+H_2O\left(l\right)\rightarrow H_3O^+\left(aq\right)+NO_3^-\left(aq\right)$ pH<7

b)
$OH\left(aq\right)+H_2O\left(l\right)\rightarrow H_2O^+\left(l\right)+OH^-\left(aq\right)$ pH>7

c)
$CH_3COOH\left(aq\right)+H_2O\left(l\right)\rightarrow H_3O^+\left(aq\right)+CH_3COO^-\left(aq\right)$ pH<7

d)
$CH_3COO^-\left(aq\right)+H_2O\left(l\right)\rightarrow CH_3COOH\left(aq\right)+OH^-\left(aq\right)$ pH>7

e)
$H_2SO_4\left(aq\right)+2H_2O\left(l\right)\rightarrow2H_3O^+\left(aq\right)+SO_4^{2-}\left(aq\right)$ pH<7

f)
$CH_3NH_2\left(aq\right)+H_2O\left(l\right)\rightarrow CH_3NH_3^+\left(aq\right)+OH^-\left(aq\right)$ pH>7

2.9
a)

$NH_3\left(aq\right)+CH_3COOH\left(aq\right)\rightarrow NH_4CH_3COO\left(aq\right)$

$NH_4OH\left(aq\right)+CH_3COOH\left(aq\right)\rightarrow NH_4CH_3COO\left(aq\right)+H_2O\left(l\right)$

Ammoniumasetaatti eli ammoniumetanaatti

n(emäs)=2,0mol

b)
$2Al\left(OH\right)_3\left(aq\right)+3H_2SO_4\left(aq\right)\rightarrow2Al\left(SO_4\right)_3+6H_2O$

Alumiinisulfaatti

n(emäs)=1,3mol

c)
$H_3PO_4\left(aq\right)+3KOH\left(aq\right)\rightarrow+K_3PO_4\left(aq\right)+3H_2O\left(l\right)$

kaliumfosfaatti

n(emäs)=2mol

d)
$2CH_3CH_2COOH\left(aq\right)+Ca\left(OH\right)_2\rightarrow Ca\left(CH_3CH_2COO\right)\left(aq\right)+2H_2O\left(l\right)$

Kalsiumpropanaatti

n(emäs)=1,0mol

2.10
a)

$c\left(H_2SO_4\right)=0{,}15\ \frac{mol}{dm^3}$

$V\left(H_2SO_4\right)=10ml=0{,}010dm^3$

$c\left(NaOH\right)=0{,}035\ \frac{mol}{dm^3}$

$V\left(NaOH\right)=?$

$2NaOH+H_2SO_4\rightarrow2H_2O+Na_2SO_4$

$n\left(H_2SO_4\right)=cV=0{,}15\ \frac{mol}{dm^3}\cdot0{,}010dm^3=0{,}0015\ mol$

$\frac{n\left(NaOH\right)}{n\left(H_2SO_4\right)}=\frac{2}{1}$ , joten

$n\left(NaOH\right)=2\cdot0{,}0015\ mol=0{,}003\ mol$

$V\left(NaOH\right)=\frac{n}{c}=\frac{0{,}003mol}{0{,}035\ \frac{mol}{dm^3}}=0{,}08571...dm^3\ \approx86ml$
b)

$m\left(HCl\right)=1{,}65g$

$M\left(HCl\right)=36{,}458\frac{g}{mol}$

$V_1\left(HCl\right)=1{,}0\ l=1{,}0dm^3$

$V_2\left(HCl\right)=40ml=0{,}040dm^3$

$c\left(Ca\left(OH\right)_2\right)=0{,}025\ \frac{mol}{dm^3}$

$V\left(Ca\left(OH\right)_2\right)=?$

$Ca\left(OH\right)_2\left(aq\right)+2HCl\left(aq\right)\rightarrow Ca\left(Cl\right)_2\left(aq\right)+2H_2O\left(l\right)$

$n\left(HCl\right)=\frac{m}{M}=\frac{1{,}65\ g}{36{,}458\ \frac{g}{mol}}=0{,}045258\ mol$

$c\left(HCl\right)=\frac{n}{V_1}=\frac{0{,}045258\ mol}{1{,}0\ dm^3}=0{,}045258\ \frac{mol}{dm^3}$

$n\left(HCl\right)=cV_2=0{,}045258\ \frac{mol}{dm^3}\cdot0{,}040\ dm^3=0{,}001812\ mol$

$\frac{n\left(Ca\left(OH\right)_2\right)}{n\left(HCl\right)}=\frac{1}{2}$ , joten $n\left(Ca\left(OH\right)_2\right)=\frac{1}{2}\cdot0{,}001812\ mol=0{,}0009060\ mol$

$V\left(Ca\left(OH\right)_2\right)=\frac{n}{c}=\frac{0{,}0009060\ mol}{0{,}025\ \frac{mol}{dm^3}}=0{,}03624\ dm^3=36\ ml$
c)

$m\left(Mg\left(OH\right)_2\right)=130\ mg=0{,}130g$

$M\left(Mg\left(OH\right)_2\right)=58{,}326\ \frac{g}{mol}$

$c\left(H_3PO_4\right)=0{,}075\ \frac{mol}{dm^3}$

$V\left(H_3PO_4\right)=?$

$2H_3PO_4\left(aq\right)+3Mg\left(OH\right)_2\left(aq\right)\rightarrow Mg_3\left(PO_4\right)_2\left(aq\right)+3H_2O\left(l\right)$

$n\left(Mg\left(OH\right)_2\right)=\frac{m}{M}=\frac{0{,}130\ g}{58{,}326\ \frac{g}{mol}}=0{,}0022289\ mol$

$\frac{n\left(H_3PO_4\right)}{n\left(Mg\left(OH\right)_2\right)}=\frac{2}{3}$ , joten $n\left(H_3PO_4\right)=\frac{2}{3}\cdot0{,}0022289\ mol=0{,}0014859\ mol$

$V\left(H_3PO_4\right)=\frac{n}{c}=\frac{0{,}0014859\ mol}{0{,}075\ \frac{mol}{dm^3}}=0{,}01981...dm^3\approx20ml$

2.11

$m\left(puhdistusaine\right)=25{,}37\ g$

$V_1\left(näyteliuos\right)=250\ cm^3=250\ ml$
$V_2\left(titrattu\ näyteliuos\right)=10{,}0\ ml$

$c\left(H_2SO_4\right)=0{,}036\ \frac{mol}{dm^3}$

$V\left(H_2SO_4\right)=37{,}3\ ml=0{,}0373dm^3$

$M\left(NH_3\right)=17{,}034\ \frac{g}{mol}$

$m-\%\left(NH_3\right)=?$

$2NH_3\left(aq\right)+H_2SO_4\left(aq\right)\rightarrow\left(NH_4\right)_2SO_4\left(aq\right)$

$n\left(H_2SO_4\right)=cV=0{,}036\ \frac{mol}{dm^3}\cdot0{,}0373\ dm^3=0{,}001343mol$

$\frac{n\left(NH_3\right)}{n\left(H_2SO_4\right)}=\frac{2}{1}$ , joten $n\left(H_3PO_4\right)=2\cdot0{,}001343\ mol=0{,}002686\ mol$

Ammoniakin massa titratussa näytteessä (10,0 ml):

$m\left(NH_3\right)=nM=0{,}002686\ mol\cdot17{,}034\frac{mol}{dm^3}=0{,}04575g$

Puhdistuainetta oli 250 ml, joten ammoniakkia olisi

$25\cdot\ 0{,}04575g=1{,}1445g$

$m-\%\left(NH_3\right)=\frac{m\left(NH_3\right)}{m\left(Näyte\right)}\cdot100\%=\frac{1{,}144g}{23{,}37g}=4{,}509\%\approx4{,}5\%$