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Question: What is the pH of \(0.1M\) \(NaHC{O_3}\)? \({K_1} = 5 \times {10^{ - 7}},{K_2} = 5 \times {10^{ - 11...

What is the pH of 0.1M0.1M NaHCO3NaHC{O_3}? K1=5×10−7,K2=5×10−11{K_1} = 5 \times {10^{ - 7}},{K_2} = 5 \times {10^{ - 11}} for carbonic acids.
A. 8.688.68
B. 9.689.68
C. 7.687.68
D. 5.585.58

Explanation

Solution

By the question we can calculate the value of pOH as the square of concentration of hydroxide ions divided by the concentration of sodium bicarbonate. And then we can calculate the value of pH by subtracting the value of pOHpOH from 1414.

Complete step by step answer:
Carbonic acid: Carbonic acid is H2CO3{H_2}C{O_3}. The value of K1{K_1}is given when one hydrogen ion is released by carbonic acid. The value of K2{K_2} is given when the second hydrogen ion is released by carbonic acid. The values of K1{K_1}and K2{K_2}, are known as dissociation constants of carbonic acid. The value for K1{K_1}is very high as compared to the value of K2{K_2}. As for K1{K_1} we have to remove hydrogen ion from stable carbonic acid.
Reaction of sodium bicarbonate is as follows:
NaHCO3+H2O→H2CO3+NaOHNaHC{O_3} + {H_2}O \to {H_2}C{O_3} + NaOH
Now, the dissociation constant for water is Kw{K_w} which is equal to 1414. The dissociation constant of carbonic acid is Ka{K_a} which is equal to the value of K1{K_1} for carbonic acid which is given in the question as 5×10−75 \times {10^{ - 7}}. Now the dissociation constant for sodium bicarbonate will be calculated using the formula Kb=KwKa{K_b} = \dfrac{{{K_w}}}{{{K_a}}}
Kb=10−145×10−7=2×10−8{K_b} = \dfrac{{{{10}^{ - 14}}}}{{5 \times {{10}^{ - 7}}}} = 2 \times {10^{ - 8}}. This is the value for the dissociation of sodium bicarbonate.
As the dissociation constant value calculated for the sodium bicarbonate is very low, so we can write
Kb=[OH]2[NaHCO3]{K_b} = \dfrac{{{{[OH]}^2}}}{{[NaHC{O_3}]}}. By using the formula we can get the concentration of hydroxide ion after putting the values of dissociation constant for sodium bicarbonate and concentration of sodium bicarbonate.
2×10−8=[OH]20.1 [OH]2=2×10−9 [OH]=4.7×10−5  2 \times {10^{ - 8}} = \dfrac{{{{[OH]}^2}}}{{0.1}} \\\ {[OH]^2} = 2 \times {10^{ - 9}} \\\ [OH] = 4.7 \times {10^{ - 5}} \\\
Now, we know that by knowing the concentration of hydroxide ion we can calculate the value of pOH as pOH is equal to the negative logarithm of the concentration of hydroxide ion.
pOH=−log⁡[OH] pOH=−log⁡(4.7×10−5) pOH=4.32  pOH = - \log [OH] \\\ pOH = - \log (4.7 \times {10^{ - 5}}) \\\ pOH = 4.32 \\\
Now by knowing the value of pOH we can calculate the value of pH as we know the relation between these two. The relation is as: pH+pOH=14pH + pOH = 14. So the value of pH will be 14−pOH14 - pOH
Hence,pH=9.68pH = 9.68.

Note:
The value of the first dissociation constant will be high than that of second dissociation constant because in the first dissociation hydrogen ion is removed from the stable compound and after first dissociation the second hydrogen ion can be removed easily because after losing that second hydrogen ion it will attain stability.