Question #210061

A weak organic acid has a Ka value of 1.2x10-6 at 25 °C.

(a) A buffer is prepared by mixing 500 mL 0.10 M of this weak acid solution with 250 mL of 0.04 M NaOH solution, which was further diluted to a volume of 1.25 L. Calculate the pH of the buffer solution.

(b) What will be the pH of this buffer solution after further addition of 250 mL of 0.02 M NaOH solution?


Expert's answer

ka=1.2×10−6k_a=1.2\times 10^{-6}


(a) V=500ml, M=0.1M ,


Ph=−log[H+]=−log(2×10−13)=13−0.301=12.7P_h=-log[H^{+}]=-log(2\times 10^{-13})=13-0.301=12.7


(b) V=250ml, M=0.02


Ph=PHa+log250500=12.7+0.301=13P_h=P_{H_a}+log\dfrac{250}{500}=12.7+0.301=13


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