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Question: Carbonic acid, \({H_2}C{O_3}\), is a diprotic acid for which \({K_1} = 4.2 \times {10^{ - 7}}\) and ...

Carbonic acid, H2CO3{H_2}C{O_3}, is a diprotic acid for which K1=4.2×10−7{K_1} = 4.2 \times {10^{ - 7}} and K2=4.7×10−11{K_2} = 4.7 \times {10^{ - 11}}. Which solution will produce a pH closest to 99?
A.0.1M  H2CO30.1M\;{H_2}C{O_3}
B.0.1M  Na2CO30.1M\;N{a_2}C{O_3}
C.0.1M  NaHCO30.1M\;NaHC{O_3}
D.0.1M  NaHCO30.1M\;NaHC{O_3} and 0.1M  Na2CO30.1M\;N{a_2}C{O_3}

Explanation

Solution

As we know that carbonic acid dissociates into hydronium ions and carbonate ions. Also the strong electrolytes are the one which completely dissociates into water almost a 100%100\% and weak electrolytes are those that dissociates less than 100%100\%

Complete step-by-step answer: Let us calculate the pH of all the given options one by one so that we can get a clear idea of pH which is close to 99.
(A)We all know that carbonic acid will dissociate as given below:
H2CO3⇌H++HCO3−{H_2}C{O_3} \rightleftharpoons {H^ + } + HCO_3^ -

In.0.10.10000
Eq.0.1−x0.1 - xxxxx

Now, we will calculate the equilibrium constant using the ratio of concentrations of reactants and products as shown below:
K1=x20.1−x ⇒4.2×10−7=x20.1 ⇒x=2.04×10−4  {K_1} = \dfrac{{{x^2}}}{{0.1 - x}} \\\ \Rightarrow 4.2 \times {10^{ - 7}} = \dfrac{{{x^2}}}{{0.1}} \\\ \Rightarrow x = 2.04 \times {10^{ - 4}} \\\
So we got the concentration of hydronium ion and we know that pH is the negative logarithm of the hydronium concentration. So the pH will be given as:
pH=−log⁡[H+] ⇒pH=−log⁡[2.04×10−4] ⇒pH=3.62  pH = - \log [{H^ + }] \\\ \Rightarrow pH = - \log [2.04 \times {10^{ - 4}}] \\\ \Rightarrow pH = 3.62 \\\
(B)Next we have sodium carbonate which on hydrolysis give bicarbonate ions and hydroxide ions as it is a strong base. We can show this reaction as:
CO32−+H2O⇌HCO3−+OH−CO_3^{2 - } + {H_2}O \rightleftharpoons HCO_3^ - + O{H^ - }
And we also know that for as strong base the formula of pH is given as:
pH=7+12pK2+12log⁡CpH = 7 + \dfrac{1}{2}p{K_2} + \dfrac{1}{2}\log C
We are given that K2=4.7×10−11{K_2} = 4.7 \times {10^{ - 11}} , therefore, the pK2=−log⁡K2p{K_2} = - \log {K_2} and it will be equivalent to:
pK2=−log⁡(4.7×10−11) ⇒pK2=10.33  p{K_2} = - \log (4.7 \times {10^{ - 11}}) \\\ \Rightarrow p{K_2} = 10.33 \\\
Thus we can calculate the value of pH as:
pH=7+12×10.33−12 ⇒pH=11.66  pH = 7 + \dfrac{1}{2} \times 10.33 - \dfrac{1}{2} \\\ \Rightarrow pH = 11.66 \\\
(C)Next we have sodium hydrogen carbonate which is a salt of strong acid and strong base, therefore the pH for this will be given by using the formula of pH as:
pH=12(pK1+pK2)pH = \dfrac{1}{2}(p{K_1} + p{K_2})
pH=12(6.38+10.33) ⇒pH=8.35  pH = \dfrac{1}{2}(6.38 + 10.33) \\\ \Rightarrow pH = 8.35 \\\
(D)Lastly we have the combination of sodium carbonate and sodium bicarbonate and their pH will be given using the formula as:
pH=pK2+log⁡[CO32−][HCO3−]pH = p{K_2} + \log \dfrac{{[CO_3^{2 - }]}}{{[HCO_3^ - ]}}
⇒pH=10.33\Rightarrow pH = 10.33
Therefore, from the above calculations we can say that the sodium bicarbonate will have the closest pH to 99.

Hence, the correct answer is option (C).

Note: Always remember that salt of strong acid and strong base are never hydrolysed because neither cation or anion of salt interact with water to produce acidity or alkalinity. Also, when weak acids and strong bases react to form a salt the anionic hydrolysis takes place and when strong acids and weak bases react, cationic hydrolysis takes place. And remember that Alkali metal hydroxides are strong bases.