Question

Determine the initial pH, the pH at the midpoint (halfway to the equivalence point), and the pH at the equivalence point if 25.00mL of 0.1335 M NH3 (NH4OH) is titrated with 0.2350M HCl? Kb = 1.75*10(-5)

Answer 1

(a). 11.2

(b). 9.24

(c). 5.16

`NH_(3(aq))+H_2O_((l))rarrNH_(4(aq))^(+)+OH_((aq))^-`

Initial moles `NH_3` = 0.1335

Initial moles `OH^-` = 0

If `x` moles of `NH_3` are used up then:

Equlibrium moles `NH_3 =( 0.1335 - x)`

Equilibrium moles `OH^-` = `x`

`K_b=([NH_4^(+)][OH^-])/([NH_3])=1.75xx10^(-5)`

Since `x` is very small we assume `(0.1335=x)rarr0.1355`

So: `([OH^-]^(2))/(0.1335)=1.75xx10^(-5)`

From which `[OH^-] = 1.53xx10^(-3)M`

`pOH=2.815`

`pH=pK_w-pOH`

`pH=14-2.815= 11.2`

At the mid - point of the titration `[NH_4^+]=NH_3]`

So `K_b=[OH^-]=1.75xx10^(-5)`

`[H^+]=(K_w)/([OH^-])=(10^(-14))/(1.75xx10^(-5))=5.714xx10^(-10)`

`pH=-log(5.714xx10^(-10))`

`pH=9.245`

We need now to get the total volume

`NH_3+HClrarrNH_4Cl+H_2O`

moles `NH_3 = (0.1335xx25)/(1000)=3.34xx10^(-3)`

So moles `HCl=3.34xx10^(-3)`

`c=n/v`

`v=n/c = (3.34xx10^(-3))/(0.235)L=14.2ml`

So total volume = `25 +14.2 = 39.2ml`

The final pH will be slightly acidic due to salt hydrolysis of `NH_4^+`:

`NH_4^(+)rightleftharpoonsNH_3+H^+`

no. moles `NH_4^(+)=3.34xx10^(-3)`

So `[NH_4^(+)]=(3.34xx10^(-3))/(39.2xx10^(-3))=0.0852M`

`K_a=([H^+]^2)/(0.0852)`

We know that `K_a=K_w/K_b=(10^(-14))/(1.75xx10^(-5))=5.7xx10^(-10)`

So `([H^+]^2)/(0.0852)=5.7xx10^(-10)`

`[H^+]=6.968xx10^(-6)`

`pH=-log(6.968xx10^(-6))`

`pH=5.16`