How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 660.0 mL of a 0.0577 M succinic acid solution to produce a pH of 5.869 ? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

the amount of grams of dipotassium succinate trihydrate  K₂C₄H₄O₄·3H₂O = 41.798g

Explanation:

Given that:

The volume of   = 660.mL

The molarity of succinic acid = 0.0577 M

The pH of the solution = 5.869

The pKa₁ = 4.207

The pKa₂ = 5.636

The pKa₂ is required to be used for the determination of given that succinic acid is dissociated twice.

Using Henderson-HasselBalch Equation,

we know that :

The volume of   = 660.mL

The molarity of succinic acid = 0.0577 M

number of moles = Molarity Volume (liters)

number of moles =  0.0577  0.660 L

number of moles =  0.038082 mol

The balanced chemical reaction for this equation can be expressed as follows:

Here, the number of moles of

number of moles of = 2 × 0.038082 mol

number of moles of = 0.076164 mol

From the Henderson-HasselBalch Equation,

SInce; the number of moles of

( 0.1302416 + 0.038082) mol

0.1683236 mol

The grams of   dipotassium succinate trihydrate = numbers of moles of dipotassium succinate trihydrate × Molar mass

The grams of   dipotassium succinate trihydrate = 0.1683236 mol × 248.32 g/mol

The grams of   dipotassium succinate trihydrate = 41.798g

Thus , the amount of grams of dipotassium succinate trihydrate  K₂C₄H₄O₄·3H₂O = 41.798g

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