Answer:
8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL
Explanation:
<em>Succinic acid has a pKa₂ of 5.63</em>
To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:
pH = pKa + log [A-] / [HA]
5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]
0.01 = log [Na₂Succ.] / [HSucc⁺]
1.0233 = [Na₂Succ.] / [HSucc⁺] <em>(1)</em>
As:
0.1M = [Na₂Succ.] + [HSucc⁺] <em>(2)</em>
Replacing (2) in (1):
1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]
1.0233[HSucc⁺] = 0.1M - [HSucc⁺]
2.033[HSucc⁺] = 0.1M
[HSucc⁺] = 0.0494M
[Na₂Succ] = 0.0506M
Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:
Na₂Succ + HCl → HSucc⁺ + Cl⁻
The moles of HCl added = Moles HSucc⁺ we need:
<em>Moles Na₂Succ:</em>
0.500L * (0.1mol/L) = 0.0500 moles
<em>Mass -Molar mass sodium succinate: 162.05g/mol-:</em>
0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added
<em>Moles HCl:</em>
0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =
And 247mL of 0.1M HCl adding enough water until make 500mL
