After the addition of NaOH to the solution, the concentration of H2CO3 is reduced to 0.095 M (0.1 - 0.005) and the concentration of HCO3^- is increased to 0.105 M (0.1 + 0.005).
We can use the k1 expression from before: k1 = (H^+)(HCO3^-)/(H2CO3)
Now we have k1 = (H^+)(0.105)/(0.095)
We already calculated k1 to be 4.2E-7, so now we can solve for the new [H^+]:
4.2E-7 = (H^+)(0.105)/(0.095)
(H^+) = 4.2E-7 * (0.095/0.105)
(H^+) = 3.8E-7
The new concentration of [H^+] after the addition of NaOH is 3.8E-7 M.
1.0L of aqueous solution in which [H2CO3]=[HCO3^-]=0.10M and has [H^+]=4.2E-7. What is the concentration of [H^+] ofter 0.005 mole of NaOH has been added?
H2CO3 ==> H^+ + HCO3^-
k1 = (H^+)(HCO3^-)/(H2CO3)
I don't know if you are supposed to calculate or to look up k1. However, you can calculate it as follows.
Since (HCO3^-) = (H2CO3) [both are 0.1 M), then plugging in 0.l M for each gives k1 = (H^+).
If we start with 0.1 mol H2CO3 and add 0.005 mol NaOH, we have 0.1 - 0.005 mols H2CO3 remaining and an extra 0.005 mol HCO3^- formed to make the final (HCO3^-)= 0.1 + 0.005 = ??
Plug k1, H2CO3, and HCO3^- into the k1 expression above and solve for (H^+). I found 3.97 x 10^-7. Post your work if you have trouble. By the way, it is a longwe way but you can also use the Henderson-Hasselbalch equation. Note also that this is a buffer solution and the H^+ didn't change much even though an amount of NaOH equivalent to 5% was added. That's what it is supposed to do. 0.005 mol NaOH added to an unbuffered solution would change the H^+ to 2 x 10^-12.
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