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A solution containing 17.22 mL of a 0.1212 M H2CO3 solution is mixed with 15.22 mL...

A solution containing 17.22 mL of a 0.1212 M H2CO3 solution is mixed with 15.22 mL of a 0.1434 M Na2CO3 solution. What is the pH of the final solution mixture? If you dilute this mixture up to 100 mL, would the pH change, and if so, what is the new pH? Ka1=6.2x10^-7 and Ka2=4.4x10^-11

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Answer #1


The solution mixture will become buffer solution,

and buffer solution pH = pKa + log([salt]/[acid])

pKa = pKa1 since Pka2 is very small value.

pH = -log(6.2*10^-7) -log((15.22*0.1434)/(17.22 * 0.1212))

pH = 6.188

-log[H+] = 6.188

[H+] = 10^-6.188

[H+] = 6.486 * 10^-7 in (17.22 + 15.22) mL solution

[H+] = 6.486 * 10^-7 M in 32.44 mL solution.

M1V1 = M2V2

6.386 * 10^-7 * 32.44 = M2 * 100

final concentration of [H+] = 2.0716 * 10^-7

pH = -log[H+]

pH = -log(2.0716 * 10^-7)

pH = 6.6836

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