Answer:
53.9 g
Explanation:
When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:
pH = pKa + log [A⁻]/[HA]
where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.
We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].
Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.
So,
4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]
⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]
0.43 = log [A⁻]/[HA]
taking antilogs to both sides of this equation:
10^0.43 = [A⁻]/[HA] = 2.69
[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M
Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:
( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻
The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):
0.34 mol x 160.21 g/mol = 53.9 g
