Question

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).

Answer 1

Answer:

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

Explanation:

Given that:

The volume of  $K_2C_4H_4O_4.3H_2O$ = 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,

$PH =pKa + log\dfrac{[Salt]}{[Acid]}$

we know that :

The volume of  $K_2C_4H_4O_4$ = 660.mL

The molarity of succinic acid = 0.0577 M

number of moles = Molarity $\times$ Volume (liters)

number of moles =  0.0577 $\times$  0.660 L

number of moles =  0.038082 mol

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

$K_2C_4H_4O_4+C_4H_4O_4^{2-} \longleftrightarrow 2KC_4H_4O_4^-$

Here, the number of moles of $K_2C_4H_4O_4= C_4H_4O_4^{2-}$

number of moles of $2KC_4H_4O_4^-$ = 2 × 0.038082 mol

number of moles of $2KC_4H_4O_4^-$ = 0.076164 mol

From the Henderson-HasselBalch Equation,

$pH =pKa + log\dfrac{[Salt]}{[Acid]}$

$pH =pKa_2 + log\dfrac{[K_2C_4H_4O_4]}{[KC_4H_4O_4^-]}$

$pH =pKa_2 + log\dfrac{n_{K_2C_4H_4O_4}}{n_{KC_4H_4O_4^-}}$

$5.869 =5.636+ log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$0.233= log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$10^{0.233}= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$1.710015315= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$1.710015315 \times 0.076164= n_{K_2C_4H_4O_4$

$n_{K_2C_4H_4O_4}= 0.1302416$

SInce; the number of moles of $K_2C_4H_4O_4= C_4H_4O_4^{2-}$

$n_{C_4H_4O_4^{2-}} =$( 0.1302416 + 0.038082) mol

$n_{C_4H_4O_4^{2-}} =$ 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