# 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)

Feb 25, 2015

(a). 11.2

(b). 9.24

(c). 5.16

$N {H}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} \rightarrow N {H}_{4 \left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

Initial moles $N {H}_{3}$ = 0.1335

Initial moles $O {H}^{-}$ = 0

If $x$ moles of $N {H}_{3}$ are used up then:

Equlibrium moles $N {H}_{3} = \left(0.1335 - x\right)$

Equilibrium moles $O {H}^{-}$ = $x$

${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \left[O {H}^{-}\right]}{\left[N {H}_{3}\right]} = 1.75 \times {10}^{- 5}$

Since $x$ is very small we assume $\left(0.1335 = x\right) \rightarrow 0.1355$

So: $\frac{{\left[O {H}^{-}\right]}^{2}}{0.1335} = 1.75 \times {10}^{- 5}$

From which $\left[O {H}^{-}\right] = 1.53 \times {10}^{- 3} M$

$p O H = 2.815$

$p H = p {K}_{w} - p O H$

$p H = 14 - 2.815 = 11.2$

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

So ${K}_{b} = \left[O {H}^{-}\right] = 1.75 \times {10}^{- 5}$

$\left[{H}^{+}\right] = \frac{{K}_{w}}{\left[O {H}^{-}\right]} = \frac{{10}^{- 14}}{1.75 \times {10}^{- 5}} = 5.714 \times {10}^{- 10}$

$p H = - \log \left(5.714 \times {10}^{- 10}\right)$

$p H = 9.245$

We need now to get the total volume

$N {H}_{3} + H C l \rightarrow N {H}_{4} C l + {H}_{2} O$

moles $N {H}_{3} = \frac{0.1335 \times 25}{1000} = 3.34 \times {10}^{- 3}$

So moles $H C l = 3.34 \times {10}^{- 3}$

$c = \frac{n}{v}$

$v = \frac{n}{c} = \frac{3.34 \times {10}^{- 3}}{0.235} L = 14.2 m l$

So total volume = $25 + 14.2 = 39.2 m l$

The final pH will be slightly acidic due to salt hydrolysis of $N {H}_{4}^{+}$:

$N {H}_{4}^{+} r i g h t \le f t h a r p \infty n s N {H}_{3} + {H}^{+}$

no. moles $N {H}_{4}^{+} = 3.34 \times {10}^{- 3}$

So $\left[N {H}_{4}^{+}\right] = \frac{3.34 \times {10}^{- 3}}{39.2 \times {10}^{- 3}} = 0.0852 M$

${K}_{a} = \frac{{\left[{H}^{+}\right]}^{2}}{0.0852}$

We know that ${K}_{a} = {K}_{w} / {K}_{b} = \frac{{10}^{- 14}}{1.75 \times {10}^{- 5}} = 5.7 \times {10}^{- 10}$

So $\frac{{\left[{H}^{+}\right]}^{2}}{0.0852} = 5.7 \times {10}^{- 10}$

$\left[{H}^{+}\right] = 6.968 \times {10}^{- 6}$

$p H = - \log \left(6.968 \times {10}^{- 6}\right)$

$p H = 5.16$