How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g

Question

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 570.0 mL of a 0.0386 M succinic acid solution to produce a pH of 5.906? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

Hint: Succinic acid (H2C4H4O4) is a diprotic acid and the succinate ion (C2H4O42–) is the completely deprotonated form of succinic acid. The addition of the succinate ion to the solution results in a buffer. Since the desired pH of the solution is greater than pKa2 for succinic acid, we know that the solution contains a mixture of the succinate ion and the monoprotonated intermediate species (HC2H4O4–).

Explanation / Answer

moles of succinic acid = molarity (M) x volume (L) = 0.0386 x 0.570 = 0.022 mols

let all of succinic acid has dissociated into monoprotonated succinate, then

moles of HC2H4O4^- = 0.022 mols

let x moles of C2H4O4^2- is present in solution

then concentration remaining of HC2H4O4^- = 0.022 - x mols

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Since pH is close to pKa2 we have,

5.906 = 5.636 + log(x/(0.022-x)

x = 1.86(0.022-x)

x = 0.041 - 1.86x

x = 0.0143 mols

So mass of dipotassium succinate trihydrate to be added to get 0.0143 mols in solution would be,

= moles x molar mass

= 0.0143 x 248.32 = 3.56 g

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