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 [Aā»]/[HA]
where Ā [Aā»] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.
We can calculate pKā from the given kā ( pKā = - log Kā ), and from there obtain the ratio Ā [Aā»]/HA].
Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 Ā can be determined.
So,
4.63 = - log ( 6.3 x 10ā»āµ ) + log [Aā»]/[HA] = - (-4.20 ) + log [Aā»]/[HA]
ā log [Aā»]/[HA] Ā = 4.63 - 4.20 = Ā log [Aā»]/[HA]
0.43 = log [Aā»]/[HA]
taking antilogs to both sides of this equation:
10^0.43 = Ā [Aā»]/[HA] = 2.69
Ā [Aā»]/ 1.00 M = 2.69 ā [Aā»] = 2.69 M
Molarity is moles per liter of solution, so we can calculate how many moles of Ā C6H5CO2ā» the student needs to dissolve Ā in 125. mL ( 0.125 L ) of a 2.69 M solution:
( 2.69 mol C6H5CO2ā» / 1L ) x 0.125 L Ā = 0.34 mol C6H5CO2ā»
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
