Question

Calculate the pH of a solution containing $0.1MHC{O}_3^{-}$ and $0.2MC{O}_3^{2-}\left[K_1\right(H_{2}C{O}_3)=4.2\times {10}^{-7}$ and $K_2\left(HC{O}_3^{-}\right)=4.8\times {10}^{-11}]$.

• A. 3.18
• B. 10.62
• C. 6.62
• D. 9.31

Answer 1

The correct option is B 10.62

$HC{O}_3^{-}+H_{2}O⟶H_3{O}^{+}+C{O}_3^{2-}{K}_2=\frac{\left[H_3{O}^{+}\right]\left[C{O}_3^{2-}\right]}{HC{O}_3^{-}}=4.8\times {10}^{-11}\left[H_3{O}^{+}\right]=4.8\times {10}^{-11}\times \frac{\left[HC{O}_3^{-}\right]}{\left[C{O}_3^{2-}\right]}=4.8\times {10}^{-11}\times \left(\frac{0.1}{0.2}\right)\Rightarrow \left[H_3{O}^{+}\right]=4.8\times {10}^{-11}\times 0.5pH=-\log \left[H_3{O}^{+}\right]=-\log [4.8\times {10}^{-11}\times 0.5]=10.62$