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

a) 129.14 g/mol

b) 8.87

Step-by-step 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 =
`(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 = (0.140M*1.00L)/(1.00L+0.116L)`

`= (0.140 mol)/(1.116 L)`

= 0.1255 M

`K_a` of HA =
`2.3*10^(-5)`

`K_b = (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 = ([HA][OH^-])/([A^-])`

`4.35*10^(-10) = ([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