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$\pu{25 mL}$ of $\pu{0.125 M}$ $\ce{Na2CO3}$ is titrated with $\pu{0.100 M}$ $\ce{HCl}$. Given that $K_\mathrm{a1} = 4.3\times 10^{-7}$ and $K_\mathrm{a2} = 4.8\times10^{-11}$ for the diprotic acid $\ce{H2CO3}$, calculate the $\mathrm{pH}$ values of the two equivalence points in the titration.

My method was to assume that all $\ce{CO3^2-}$ initially would be neutralized to form $\ce{HCO3-}$, and from there use the $K_\mathrm{a2}$ value to calculate the concentration of $\ce{H3O+}$ via equilibrium, but this gives me the wrong answer (by about $3$ $\mathrm{pH}$ units).

Could someone help me understand what I'm doing wrong, and how I should do it correctly? (The correct answer apparently should be $8.34$ for the 1st equivalence point.)

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  • $\begingroup$ Use $K_\mathrm{b1}=\frac{K_\mathrm{w}}{K_\mathrm{a1}}$ in addition to $K_\mathrm{a2}$ in order to calculate $[\ce{H3O+}]$ $\endgroup$ – Adnan AL-Amleh Jan 28 at 18:56
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I think the main confusion you're having here is that you misinterpretted the composition of the system at the first end point.

While you are correct that an equivalent non-equilibrium system is just all $\ce{HCO3-}$, it is not correct to think of this as an acid. This is bicarbonate. If you take sodium bicarbonate and dissolve it water, it is basic, not acidic. Therefore, you should have converted your $K_{\mathrm{a1}}$ value into a $K_{\mathrm{b}}$ value and treated the system as a base instead.

The second equivalence point is when you have the equivalent concentration of $\ce{H2CO3}$. Treat this as a normal acid dissociation problem.

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At the first equivalence point: $$\ce{Na_2CO_3 +HCl ->NaHCO_3 + NaCl}$$ $$\text{moles of}~ \ce{Na_2CO_3} =25\times{0.125}\times{10^{-3}}=3.125\times{10^{-3}}\pu{mol}=\text{moles of}~ \ce{HCl}=\text{moles of}~\ce{NaHCO3} $$ $$\text{moles of}~ \ce{HCl} =3.125\times{10^{-3}}=\pu{V_\ce{HCl}\times{0.1}} $$ $$\pu{V_\ce{HCl}}=31.25\times{10^{-3}}$$ $$\text{Total volume}=25\times{10^{-3}}+31.25\times{10^{-3}}=56.25\times{10^{-3}}~\pu{L}$$ $$[\ce{NaHCO3}]=\frac{3.125\times{10^{-3}}\pu{mol}}{56.25\times{10^{-3}}~\pu{L}}=5.6\times{10^{-2}}\pu{M}=\ce[{HCO_3^-}]$$ $$[\ce{H_3O^+}]=\sqrt\frac{K_\mathrm{a1}K_\mathrm{a2}[\ce{NaHCO_3}]+K_\mathrm{a1}K_\mathrm{w}}{K_\mathrm{a1}+[\ce{NaHCO_3}]}$$

$$[\ce{H_3O^+}]=\sqrt\frac{(4\cdot{3}\times{10^{-7}}\times{4\cdot{8}\times{10^{-11}}}\times{0.056})+(4\cdot{3}\times{10^{-7}}\times{10^{-14}})}{(4\cdot{3}\times{10^{-7}}+0.056)}\approx{5.35\times{10^{-9}}}\pu~{M}$$ $$\pu{pH}={8.27}\approx{8.3}$$

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At the second equivalence point: $$\ce{Na_2CO_3 +2HCl ->H_2CO_3 + 2NaCl}$$ $$\text{moles of}~ \ce{Na_2CO_3} =25\times{0.125}\times{10^{-3}}=3.125\times{10^{-3}}\pu{mol}=\text{moles of}~\ce{H_2CO3} $$ $$\text{moles of}~ \ce{HCl} =2\times{3.125\times{10^{-3}}}=\pu{V_\ce{HCl}\times{0.1}} $$ $$\pu{V_\ce{HCl}}=62.25\times{10^{-3}}$$ $$\text{Total volume}=25\times{10^{-3}}+62.25\times{10^{-3}}=87.25\times{10^{-3}}~\pu{L}$$ $$[\ce{H_2CO3}]=\frac{3.125\times{10^{-3}}\pu{mol}}{87.25\times{10^{-3}}~\pu{L}}=3.58\times{10^{-2}}\pu{M}$$ Treat carbonic acid solutions as if $\ce{H2CO3}$ were monoprotic, because the first acid dissociation constant is much greater than either$ K_\mathrm{a2}$ or $ K_\mathrm{w}$ , so this becomes a standard monoprotic weak acid problem:

$$ K_\mathrm{a1} = 4.3\times{10^{–7}} =\frac{[\ce{H^+}][\ce{HCO_3^− }]}{[\ce{H_2CO_3}]}$$ Assume : $ [\ce{H+}] = [\ce{HCO_3^-} ]$ , the equilibrium expression becomes: $$K_\mathrm{a1} = 4.3\times{10^{–7}} =\frac{[\ce{H^+}]^2}{0.0358-[\ce{H+}] }$$ As: $0.0358 >> K_\mathrm{a1}$,justifies the further approximation of dropping the $ [\ce{H+}]$ term in the denominator. $$ K_\mathrm{a1} =4.3\times{10^{-7}} =\frac{[\ce{H^+}]^2}{0.0358}$$ $$[\ce{H^+}]=1.24\times{10^{-4}}\pu{M}$$ $$\pu{pH}=3.92$$

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$pK_{a1} = -\log{(4.3\times10^{-7})} = 6.3665$

$pK_{a2} = -\log{(4.8\times10^{-11})} = 10.3188$

eq. pt. $ = \dfrac{6.3665+10.3188}{2} = 8.34265 \ce{->[Rounding]} 8.34$

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  • $\begingroup$ but why does this work? also this cant be extended to calculate the second equivalence point $\endgroup$ – ゴトひめ Sep 20 '18 at 15:28
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    $\begingroup$ At eq. pt $\ce{[H2CO3] = [CO3^{2-}]}$. // What would you average with 6.3665 to get second eq. pt?!? $\endgroup$ – MaxW Sep 20 '18 at 15:32
  • $\begingroup$ This relationship is also reasonably easy to derive using the corresponding Henderson-Hasselbalch equations, if you've learned though already. $\endgroup$ – Zhe Sep 20 '18 at 15:42
  • $\begingroup$ At second eq. pt. (for $K_{a1}$) $\ce{[H+] = [HCO3-]}$. I'll leave the math to you. $\endgroup$ – MaxW Sep 20 '18 at 15:56
  • $\begingroup$ $$\pu{pH}=\frac{1}{2}\times{(pK_\mathrm{a1}+pK_\mathrm{a2} )}$$ THis foumula suitable when $K_\mathrm{a1}<<[\ce{HCO_3^-}]\text{with concentrated solution,but neglegt water autoionization and assume} [\ce{H2CO3}] = [\ce{CO3-} ]$ $\endgroup$ – Adnan AL-Amleh Jan 27 at 21:44

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