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
[tex]V_{equivalence}_{acid}[/tex] = 1.00 L
Equation for the reaction can be represented as:
[tex]NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)[/tex]
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 = [tex]\frac{18.08g}{0.140 mole}[/tex]
= 129.14 g/mol
b) since pH = pKa
Then pKa of HA = 4.63
Ka = [tex]10^{-4.63][/tex]
= [tex]2.3*10^{-5}[/tex]
[tex][A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}[/tex]
[tex]= \frac{0.140 mol}{1.116 L}[/tex]
= 0.1255 M
[tex]K_a[/tex] of HA = [tex]2.3*10^{-5}[/tex]
[tex]K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}[/tex]
[tex]= 4.35*10^{-10}[/tex]
[tex]A_{(aq)}[/tex] + [tex]H_2O_{(l)}[/tex] [tex]\rightleftharpoons[/tex] [tex]HA_{(aq)}[/tex] + [tex]OH^-_{(aq)}[/tex]
Initial 0.1255 0 0
Change - x + x + x
Equilibrium 0.1255 - x x x
[tex]K_b = \frac{[HA][OH^-]}{[A^-]}[/tex]
[tex]4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}[/tex]
As [tex]K_b[/tex] is very small, (o.1255 - x) = 0.1255
[tex]x = \sqrt{0.1255*4.35*10^{-10}}[/tex]
[OH⁻] = [tex]x = 7.4 *10^{-6}[/tex]
But pOH = - log [OH⁻]
= - log [[tex]7.4*10^{-6}[/tex]]
= 5.13
pH = 14.00 = 5.13
pH = 8.87
