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Question: Citric acid \[\left( {{H_3}A} \right)\] is a polyprotic acid with \[{K_1},{K_2}\] and \({K_3}\) equa...

Citric acid (H3A)\left( {{H_3}A} \right) is a polyprotic acid with K1,K2{K_1},{K_2} and K3{K_3} equals to 7.4×10−4,1.7×10−57.4 \times {10^{ - 4}},1.7 \times {10^{ - 5}} and 4.0×10−74.0 \times {10^{ - 7}} ,respectively. Calculate the [H⊕],[H2A−],[HA2−][{H^ \oplus }],\left[ {{H_2}{A^ - }} \right],\left[ {H{A^{2 - }}} \right] and [A3−]\left[ {{A^{3 - }}} \right] in 0.01M0.01M citric acid.
A. [H⊕]=[H2A−]=2.4×10−3M,[HA2−]=1.7×10−5,[A3−]=2.8×10−9\left[ {{H^ \oplus }} \right] = \left[ {{H_2}{A^ - }} \right] = 2.4 \times {10^{ - 3}}M,\left[ {H{A^{2 - }}} \right] = 1.7 \times {10^{ - 5}},\left[ {{A^{3 - }}} \right] = 2.8 \times {10^{ - 9}}
B. [H⊕]=[H2A−]=2.4×10−6M,[HA2−]=1.75×10−5,[A3−]=2.8×10−6\left[ {{H^ \oplus }} \right] = \left[ {{H_2}{A^ - }} \right] = 2.4 \times {10^{ - 6}}M,\left[ {H{A^{2 - }}} \right] = 1.75 \times {10^{ - 5}},\left[ {{A^{3 - }}} \right] = 2.8 \times {10^{ - 6}}
C. [H⊕]=[H2A−]=2.4×10−5M,[HA2−]=1.75×10−5,[A3−]=2.8×10−8\left[ {{H^ \oplus }} \right] = \left[ {{H_2}{A^ - }} \right] = 2.4 \times {10^{ - 5}}M,\left[ {H{A^{2 - }}} \right] = 1.75 \times {10^{ - 5}},\left[ {{A^{3 - }}} \right] = 2.8 \times {10^{ - 8}}
D. [H⊕]=[H2A−]=2.4×10−7M,[HA2−]=1.75×10−5,[A3−]=2.8×10−6\left[ {{H^ \oplus }} \right] = \left[ {{H_2}{A^ - }} \right] = 2.4 \times {10^{ - 7}}M,\left[ {H{A^{2 - }}} \right] = 1.75 \times {10^{ - 5}},\left[ {{A^{3 - }}} \right] = 2.8 \times {10^{ - 6}}

Explanation

Solution

We know that Citric acid (H3A)\left( {{H_3}A} \right) is a polyprotic acid, polyprotic acids contain multiple acidic protons and dissociates in different steps. In polyprotic acid the first dissociation constant is always the largest followed by second. This is because the protons become successively less acidic as they are lost.

Complete step by step answer:

To solve this problem first of all Let's write the dissociation equation of citric acid which is as follows;
H3A+H2O⇌H2A−+H3O⊕{H_3}A + {H_2}O \rightleftharpoons {H_2}{A^ - } + {H_3}{O^ \oplus } (First dissociation reaction)
So the value of first dissociation constant will be the ratio of the concentrations of product and reactant; K1=[H2A−][H3O⊕][H3A]=7.4×10−4{K_1} = \dfrac{{\left[ {{H_2}{A^ - }} \right]\left[ {{H_3}{O^ \oplus }} \right]}}{{[{H_3}A]}} = 7.4 \times {10^{ - 4}}
Let suppose the amount of [H2A−],[H3O⊕]\left[ {{H_2}{A^ - }} \right],\left[ {{H_3}{O^ \oplus }} \right]formed is xMxM of each after dissociation of citric acid then, K1=x20.01−x=7.4×10−4{K_1} = \dfrac{{{x^2}}}{{0.01 - x}} = 7.4 \times {10^{ - 4}}
⇒x2+(7.4×10−4)x−(7.4×10−6)=0 ⇒x=2.4×10−3M \begin{gathered} \Rightarrow {x^2} + \left( {7.4 \times {{10}^{ - 4}}} \right)x - \left( {7.4 \times {{10}^{ - 6}}} \right) = 0 \\\ \Rightarrow x = 2.4 \times {10^{ - 3}}M \\\ \end{gathered}
Hence the concentration of [H3O⊕]\left[ {{H_3}{O^ \oplus }} \right] and [H2A−]\left[ {{H_2}{A^ - }} \right] is 2.4×10−3M2.4 \times {10^{ - 3}}M.
Now H2A−{H_2}{A^ - } will further dissociate in HA2−H{A^{2 - }} and H3O⊕{H_3}{O^ \oplus }. The dissociation equation of H2A−{H_2}{A^ - } is as follows; H2A−+H2O⇌HA2−+H3O⊕{H_2}{A^ - } + {H_2}O \rightleftharpoons H{A^{2 - }} + {H_3}{O^ \oplus } (Second dissociation reaction)
So the second dissociation constant will be;
K2=[HA2−][H3O⊕][H2A−]{K_2} = \dfrac{{\left[ {H{A^{2 - }}} \right]\left[ {{H_3}{O^ \oplus }} \right]}}{{\left[ {{H_2}{A^ - }} \right]}} ; let the concentration of HA2−H{A^{2 - }} and H3O⊕{H_3}{O^ \oplus } formed be yMyM.
⇒1.7×10−5=[HA2−][y]y ⇒[HA2−]=1.7×10−5M \begin{gathered} \Rightarrow 1.7 \times {10^{ - 5}} = \dfrac{{\left[ {H{A^{2 - }}} \right]\left[ y \right]}}{y} \\\ \Rightarrow \left[ {H{A^{2 - }}} \right] = 1.7 \times {10^{ - 5}}M \\\ \end{gathered}
HA2−H{A^{2 - }} will further dissociate in A3−{A^{3 - }} and H3O⊕{H_3}{O^ \oplus }. The dissociation equation of HA2−H{A^{2 - }} is given below;
HA2−+H2O⇌A3−+H3O⊕H{A^{2 - }} + {H_2}O \rightleftharpoons {A^{3 - }} + {H_3}{O^ \oplus }; (Third dissociation reaction)
Hence the value of the third dissociation constant will be ; K3=[A3−][H3O⊕][HA2−]{K_3} = \dfrac{{\left[ {{A^{3 - }}} \right]\left[ {{H_3}{O^ \oplus }} \right]}}{{\left[ {H{A^{2 - }}} \right]}}
Here we can directly put the concentration of A3−{A^{3 - }} and H3O⊕{H_3}{O^ \oplus } in the equation of the third dissociation constant as we have calculated it above.
Now; ⇒4.0×10−7=[A3−](2.4×10−3)1.7×10−5 ⇒[A3−]=2.8×10−9M \begin{gathered} \Rightarrow 4.0 \times {10^{ - 7}} = \dfrac{{\left[ {{A^{3 - }}} \right]\left( {2.4 \times {{10}^{ - 3}}} \right)}}{{1.7 \times {{10}^{ - 5}}}} \\\ \Rightarrow \left[ {{A^{3 - }}} \right] = 2.8 \times {10^{ - 9}}M \\\ \end{gathered}
Hence we have calculated[H⊕]=[H2A−]=2.4×10−3M,[HA2−]=1.7×10−5,[A3−]=2.8×10−9\left[ {{H^ \oplus }} \right] = \left[ {{H_2}{A^ - }} \right] = 2.4 \times {10^{ - 3}}M,\left[ {H{A^{2 - }}} \right] = 1.7 \times {10^{ - 5}},\left[ {{A^{3 - }}} \right] = 2.8 \times {10^{ - 9}}

So, the correct answer is “Option A”.

Note:
1- We know that in the problem statement the value of first, second and third dissociation constant of citric aid is given.
2- So with the help of dissociation constant and dissociation equation we have calculated the value of concentration of the product.