# Question 45fc9

Nov 24, 2016

$\textsf{p H = 4.88}$

#### Explanation:

Phenylamine (aniline) is a weak base and will react with the acid on mixing:

$\textsf{{C}_{6} {H}_{5} N {H}_{2} + {H}^{+} \rightarrow {C}_{6} {H}_{5} N {H}_{3}^{+}}$

The no. moles acid added is given by:

$\textsf{{n}_{{H}^{+}} = c \times v = 0.15 \times \frac{100}{1000} = 0.015}$

The no. moles phenylamine is given by:

$\textsf{{n}_{{C}_{6} {H}_{5} N {H}_{2}} = c \times v = 0.2 \times \frac{200}{1000} = 0.04}$

You can see that the no. moles of phenylamine is INXS so the number remaining after reaction $\textsf{= 0.04 - 0.015 = 0.025}$

From the equation you can see that the no. moles of $\textsf{{C}_{6} {H}_{5} N {H}_{3}^{+}}$ formed =$\textsf{0.015}$.

We have created a buffer solution containing a weak base and its salt.

The route to the solution is to find $\textsf{\left[O {H}^{-}\right] \rightarrow \left[{H}^{+}\right] \rightarrow p H}$

Phenylamine is a weak base and dissociates in water:

$\textsf{{C}_{6} {H}_{5} N {H}_{2} + {H}_{2} O r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{5} N {H}_{3}^{+} + O {H}^{-}}$

For which:

$\textsf{{K}_{b} = \frac{\left[{C}_{6} {H}_{5} N {H}_{3}^{+}\right] \left[O {H}^{-}\right]}{\left[{C}_{6} {H}_{5} N {H}_{2}\right]} = 4.6 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

These are equilibrium concentrations.

Rearranging:

sf([OH^-]=K_bxx([C_6H_5NH_2])/([C_6H_5NH_3^+])#

We don't actually need the total volume to find the concentrations as this is common to base and salt so cancels.

This means we can use the mole values which have been calculated. We assume that, because the dissociation is small, these values will approximate to equilibrium values.

Putting in the numbers:

$\textsf{\left[O {H}^{-}\right] = 4.6 \times {10}^{- 10} \times \frac{0.025}{0.015} = 7.66 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p O H = - \log \left[O {H}^{-}\right] = - \log \left(7.66 \times {10}^{- 10}\right) = 9.115}$

$\textsf{p H + p O H = 14}$

$\therefore$$\textsf{p H = 14 - p O H = 14 - 9.115 = 4.88}$