23,

`HNO_2` ⇌ `H^+` + `NO_2^-`

pH = 3,2 `to` `[H^+]` = `[NO_2^-]` = `10^(-3.2)`

`[HNO_2]` = `c_s-[H^+]`

`K=([NO_2^-]*[H^+])/([HNO_2])`

`4*10^(-4)` = `(10^(-3.2))^2/(c_s-10^(-3.2))`

`c_s` = `ul(1.63*10^(-3))` `(mol)/(dm^3)`.

24,

`M_(NH_3)` = 17,03 `g/(mol)`

`NH_3` + `H_2O` ⇌ `NH_4^+` + `OH^-`

`c_b` = ?

pH = 11 `to` `pOH` = 14-11 = 3 `to` `[OH^-]` = `10^(-3)` `(mol)/(dm^3)` = `[NH_4^+]`

`K_b` = `([NH_4^+]*[OH^-])/([NH_3])`

`1.79*10^(-5)=(10^(-3))^2/(c_b-10^(-3))`

`c_b` = `5.69*10^(-2)` `(mol)/(dm^3)`

500 `cm^3` oldatban van feleannyi, vagyis `2.84*10^(-2)` mol ammónia, ami `2.84*10^(-2)*17.03` = 0,484 g ammónia

Az eredeti oldat

100 g-jában van 25 g ammónia

hány g oldatban van 0,484 g ammónia?

m = `0.484/25*100` = 1,94 g

V = `m/rho` = `1.94/0.892` = `ul("2,17 " cm^3)` a kiindulási ammónia-oldat.

25,

`M_(CH_3COOH)` = 60,06 `g/(mol)`

`CH_3COOH` ⇌ `CH_3COO^-` + `H^+`

pH = 4,4 `to` `[H^+]` = `10^(-4.4)` = `[CH_3COO^-]`

`[CH_3COOH]` = `c_s-[H^+]`

`K_s=([CH_3COO^-]*[H^+])/([CH_3COOH])`

`1.8*10^(-5)=(10^(-4.4))^2/(c_s-10^(-4.4))`

`c_s` = `1.28*10^(-4)` `(mol)/(dm^3)`

m = `c_s*M` = `1.28*10^(-4)*60.06` = `7.6*10^(-3)` g

V = `m/rho` = `(7.6*10^(-3))/1.06` = `ul(7.24*10^(-3))` `cm^3` jégecetre van szükség.

26,

`M_(CH_3COOH)` = 60,06 `g/(mol)`

`CH_3COOH` ⇌ `CH_3COO^-` + `H^+`

A készített oldat:

pH = 4 `to` `[H^+]` = `10^(-4)` = `[CH_3COO^-]`

`[CH_3COOH]` = `c_s-[H^+]`

`K_s=([CH_3COO^-]*[H^+])/([CH_3COOH])`

`1.8*10^(-5)=(10^(-4))^2/(c_s-10^(-4))`

`c_s` = `6.56*10^(-4)` `(mol)/(dm^3)`-es.

Az eredeti oldat:

100 g-jában van 42 g ecetsav

`100/1.05` `cm^3`-ében van `42/60.06` mol ecetsav

1000 `cm^3`-ében van `1000/(100/1.05)*(42/60.06)` = 7,34 mol ecetsav

A hígítás tehát:

`7.34/(6.56*10^(-4))` `approx` 11000-szeres.

27,

`M_(NH_3)` = 17,03 `g/(mol)`

A kiindulási oldat

100 g-jában van 20 g ammónia

`100/0.91` `cm^3` oldatban van `20/17.03` mol ammónia

ezt hússzorosára hígítva:

`20*1000/0.91` `cm^3` oldatban van `20/17.03` mol ammónia

1000 `cm^3` oldatban van `1000/(20*1000/0.91)*(20/17.03)` = 0,053 `(mol)/(dm^3)`-es a hígított ammóniaoldatunk.

`c_b` = 0,0534

`NH_3` + `H_2O` ⇌ `NH_4^+` + `OH^-`

`K_b=([NH_4^+]*[OH^-])/([NH_3])`

`1.79*10^(-5)=x^2/(0.0534-x)`

x = `9.69*10^(-4)` = `[OH^-]`

pOH = `-lg[OH^-]` = `-lg(9.69*10^(-4))` = 3,01

pH = 14-pOH = 14-3,01 = `ul("10,99")`.