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Suppose there are $\pu{0.024}$ moles of $\ce{Na2HPO4}$ in a $\pu{0.250L}$ solution. How would we calculate the amount of $\pu{0.145M}$ $\ce{HCl}$ needed to reach a $\mathrm{pH}$ of 6.60?

I have attempted to solve the problem using the following steps:

  1. The final solution needs a proton concentration of $\pu{10^{-6.6} moles/L}$

  2. However when we add $\ce{HCl}$, the base would react to produce phosphoric acid and $\ce{NaCl}$. $\ce{Na2HPO4 + HCl -> H3PO4 + NaCl}$

Thus the amount of $\ce{HCl}$ needed to neutralize the weak base is equal to the moles of the base which is 0.024.

Afterwards, I attempted to find the volume of $\ce{HCl}$ needed to be added taking the $\pu{0.024 moles}$ needed into account. However, I can’t figure out how the produced phosphoric acid or amphiprotic $\ce{HPO4^2-}$ contributes to the $\mathrm{pH}$ quantitatively as no $K_\mathrm{a}/K_\mathrm{b}$ value is given in the question.

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    $\begingroup$ Welcome to Chemistry.SE. We’d like you to take the Tour to get familiarized with the site and our policy towards homework questions. I'd like to point out that your chemical equation is incorrect (step 2). Keep also in mind that the acid has three $K_\mathrm{a}$ values and you need $K_\mathrm{a2}$ to resolve this question. $\endgroup$ – Mathew Mahindaratne May 29 '18 at 2:17
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For this you have to know the $pK_{a_2}$ of $\ce{H_3PO4}$.Experimental data shows, $pK_{a_2} = 7.2$.
Now suppose you have added $x$ L. of $\ce{HCl}$. Now the reaction which you have written will happen when you add $\ce{HCl}$ in excess. Initially the following reaction will happen $$\ce{Na2HPO4 + HCl -> NaH2PO4 +_ NaCl}$$ if we add acid quantitatively. So, moles of $\ce{HCl}$ added will be $x \times 0.145$, and that's why if we add the acid in such a way that the whole acid gets consumed; after the whole reaction, remaining moles of $\ce{Na2HPO4 }$ will be $0.024 - (x \times 0.145)$, and the moles of $\ce{NaH2PO4}$ produced will be same as the acid consumed.
Now the solution contains a mixture of $\ce{HPO4^2-}$ and $\ce{H2PO4-}$, which acts as a acid buffer. Applying the Henderson-Haselbach equation we will have, $$pH = pK_a + log\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}$$ which gives us $6.6 = 7.2 + log\frac{0.024- (x \times 0.145)}{x \times 0.145}$
Solving you will get $x =0.132 L.$ So, you need to add $0.132 $ litres of acid to get that $pH$ of 6.6.

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  • $\begingroup$ Thank you - the question did not have the acid dissociation constant and I now know that it expects us to search it. $\endgroup$ – Dogukan Kayhan May 29 '18 at 14:36

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