Question

My answer(s) are incorrect. Please help. The Ka values for H2CO3 are: Ka1=4.45x10^-7 and Ka2=4.69x10^-11.

How would you prepare 1.20 L of a buffer with a pH of 9.13 from 0.270 M Na2CO3 and 0.190 M HCl? (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) Mix 427 mL of 0.270 M Na2CO3 with 773 mL of 0.190 M HCI. An error has been detected in your answer. Check for typos, miscalculations etc. before submitting your answer.

Answer 1

The reaction between Na2CO3 and HCl: Na2CO3 + HCl ----> NaHCO3 + NaCl

pKa2 of NaCO3 = -Log(Ka2) = -Log(4.69x10^-11) = 10.33

According to the Henderson-Hasselbulch equation:

pH = pKa + Log([Na2CO3]/[NaHCO3])

i.e. 9.13 = 10.33 + Log([Na2CO3]/[NaHCO3])

i.e. Log([Na2CO3]/[NaHCO3]) = -1.2

i.e. [Na2CO3]/[NaHCO3] = 10^-1.2 = 0.0631

Now, the pecent fraction of Na2CO3 = {0.0631/(1+0.0631)}*100 = 5.9%

And the percent fraction of NaHCO3 = 100-5.9 = 94.1%

Accordingly, you can add HCl to the Na2CO3 solution.