Answer:
9.744g of monosodium succinate.
4.925g of disodium succinate.
Explanation:
To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:
pH = pKa + log ([Na₂Suc] / [NaHSuc])
As you want a pH of 5.28 and pKa is 5.64:
5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])
-0.36 = log ([Na₂Suc] / [NaHSuc])
0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>
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As total concentration of the buffer is 100mM = 0.100M:
0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>
Replacing (2) in (1):
0.4365 = (0.100M - [NaHSuc] / [NaHSuc])
0.4365 = (0.100M - [NaHSuc] / [NaHSuc])
0.4365 [NaHSuc] = 0.100M - [NaHSuc]
1.4365 [NaHSuc] = 0.100M
[NaHSuc] = 0.0696M
And:
[Na₂Suc] = 0.0304M
As volume of the buffer is 1L:
[NaHSuc] = 0.0696 moles
[Na₂Suc] = 0.0304 moles
Using molar mass of both substances:
Mass of monosodium succinate:
0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>
Mass of disodium succinate:
0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>
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