How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

Succinic acid has a pKa₂ of 5.63

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⁺] (1)

As:

0.1M = [Na₂Succ.] + [HSucc⁺] (2)

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:

Moles Na₂Succ:

0.500L * (0.1mol/L) = 0.0500 moles

Mass -Molar mass sodium succinate: 162.05g/mol-:

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

Moles HCl:

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

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