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Question

Chemistry Question on Equilibrium

What is the pHpH of 0.01 M0.01\, M glycine solution? For glycine Ka1=4.5×10−3K_{a_1} = 4.5 \times 10^{-3} and Ka2=1.7×10−10K_{a_2} = 1.7 \times 10^{-10} at 298 K298 \,K.

A

3

B

10

C

44743

D

8.2

Answer

44743

Explanation

Solution

K=Ka1×Ka2=4.5×10−3×1.7×10−10 K=K_{a_{1}} \times K_{a_{2}} =4.5 \times 10^{-3} \times 1.7 \times 10^{-10} =7.65×10−13=7.65 \times 10^{-13} [H+]=K⋅C\left[ H ^{+}\right] =\sqrt{K \cdot C} =7.65×10−13×0.01=\sqrt{7.65 \times 10^{-13} \times 0.01} =8.7×10−8=8.7 \times 10^{-8} pH=−log⁡[H+]=−log⁡8.7×10−8pH =-\log \left[ H ^{+}\right]=-\log 8.7 \times 10^{-8} =−(log⁡10−8+log⁡8.7)=-\left(\log 10^{-8}+\log 8.7\right) =8−0.93=8-0.93 ⇒=7.07 \Rightarrow =7.07