Question

A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of solution and was then titrated with 0.140 M . After 500.0 mL was added, the pH was 4.63. The experimenter found that 1.00 L of 0.140 M was required to reach the stoichiometric point of the titration. a What is the molar mass of ? Molar mass = 129.14 g/mol b Calculate the pH of the solution at the stoichiometric point of the titration.

Answer 1

Answer:

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

$V_{equivalence}_{acid}$ = 1.00 L

Equation for the reaction can be represented as:

$NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)$

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = $\frac{18.08g}{0.140 mole}$

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = $10^{-4.63]$

= $2.3*10^{-5}$

$[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}$

$= \frac{0.140 mol}{1.116 L}$

= 0.1255 M

$K_a$ of HA = $2.3*10^{-5}$

$K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}$

$= 4.35*10^{-10}$

                     $A_{(aq)}$     +     $H_2O_{(l)}$         $\rightleftharpoons$     $HA_{(aq)}$     +     $OH^-_{(aq)}$

Initial        0.1255                                            0                    0

Change     - x                                                  +  x                 + x

Equilibrium   0.1255 - x                                   x                    x

$K_b = \frac{[HA][OH^-]}{[A^-]}$

$4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}$

As $K_b$ is very small, (o.1255 - x) = 0.1255

$x = \sqrt{0.1255*4.35*10^{-10}}$

[OH⁻] = $x = 7.4 *10^{-6}$

But pOH = - log [OH⁻]

= - log [$7.4*10^{-6}$]

= 5.13

pH = 14.00 = 5.13

pH = 8.87