a solution containing 2733 ml of a 01234 M Na2CO3 solution i

a solution containing 27.33 ml of a 0.1234 M Na2CO3 solution is mixed with 22.33ml of a 0.1212M H2CO3. What is the pH of the final solution mixture? KA1=6.2x10^7 KA2=4.4x-^-11

b. if you diluted to 100ml would the pH change and why?

c. what would the new pH be?

Solution

moles of Na2CO3 = 0.02733*0.1234 = 3.37*10^-3 moles

moles of H2CO3 = 0.02233*0.1212 = 2.71*10^-3 moles

H2CO3 -----> H+ + HCO3-

2.71*10^-3-x x x

ka1 = 6.2*10^-7 = X^2/(2.71*10^-3-x)

x= 5.76*10^-5 moles = moles of HCO3-

HCO3- -----> H+ + CO32-

5.76*10^-5-x 5.76*10^-5+x x

ka2 = (5.76*10^-5+x)*x/(5.76*10^-5-x) = 4.4*10^-11

x= 4.399*10^-11

so moles of H+ = 5.76*10^-5

[CO3 2- ] = 4.4*10^-11/0.04966= 8.86*10^-10M

[H2CO3] = 2.71*10^-3-5.76*10^-5 / (0.04966) =0.0534M

2pH = pKA1 + pKA2 + log10([CO3 2-]/[H2CO3])

pH = 4.39......................................................................ANS

if solution is diluted to 100 ml

the pH will decrease little as on dilution ka1 and ka2 will increase