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Question: What is the pH of \(0.01{\text{ M}}\) glycine solution? [For glycine, \({K_{{a_1}}} = 4.5 \times {...

What is the pH of 0.01 M0.01{\text{ M}} glycine solution?
[For glycine, Ka1=4.5×10−3{K_{{a_1}}} = 4.5 \times {10^{ - 3}}and Ka2=1.72×10−10{K_{{a_2}}} = 1.72 \times {10^{ - 10}}at 298K298K]
A.3.03.0
B.10.010.0
C.6.16.1
D.7.17.1

Explanation

Solution

To answer this question, you must recall the formula for dissociation constant. For a two-step dissociation, the dissociation constant for the complete reaction is calculated by the multiplication of the dissociation constants of the first and second step. Then, we shall substitute the value in the given equation and calculate the pH from there.
Formula Used : [H+]=K×c\left[ {{H^ + }} \right] = \sqrt {K \times c}
Where K is dissociation constant and c in the concentration.

Complete step by step answer:
We are the given the values of the two dissociation constants as Ka1=4.5×10−3{K_{{a_1}}} = 4.5 \times {10^{ - 3}}and Ka2=1.72×10−10{K_{{a_2}}} = 1.72 \times {10^{ - 10}}
The dissociation constant for the overall process can be calculated by multiplying the two values and we get, K=Ka1×Ka2=4.5×10−3×1.72×10−10K = {K_{{a_1}}} \times {K_{{a_2}}} = 4.5 \times {10^{ - 3}} \times 1.72 \times {10^{ - 10}}
K=7.65×10−13K = 7.65 \times {10^{ - 13}}
We know the relation to find the concentration of hydrogen ions in the solutions, given the concentration of the acid and its dissociation constant as [H+]=K×c\left[ {{H^ + }} \right] = \sqrt {K \times c}
Substituting the values of dissociation constant and concentration of the acid in the formula, we get,
[H+]=7.65×10−13×0.01M\left[ {{H^ + }} \right] = \sqrt {7.65 \times {{10}^{ - 13}} \times 0.01{\text{M}}}
[H+]=8.74×10−8M\left[ {{H^ + }} \right] = 8.74 \times {10^{ - 8}}{\text{M}}
Now that we have the hydrogen ion concentration, we can find the pH of the solution easily by finding the negative log of the hydrogen ion concentration in the glycine solution. We get,
pH=−log⁡[H+]pH = - \log \left[ {{{\text{H}}^ + }} \right]
Substituting the values:
⇒pH=−log⁡(8.74×10−8M)\Rightarrow pH = - \log \left( {8.74 \times {{10}^{ - 8}}{\text{M}}} \right)
∴pH=7.06\therefore pH = 7.06
Rounding off the value to the nearest one tenths place we can say that pH=7.1pH = 7.1

Thus, the correct answer is D.

Note:
The pH of any solution depends directly on the hydrogen ion concentration or the hydroxide ion concentration in the solution. The concentration of the hydrogen ions in a solution is inversely proportional to that of hydroxide ions. If the concentration of hydrogen ions in a solution increases, the solution becomes more acidic and if the concentration of hydroxide ions in a solution increases, the solution becomes more alkaline.
Greater is the hydrogen ion concentration, smaller is the magnitude of pH and more acidic is the solution.