Question

Rew Topics Rolorences Use the References to access important values if needed for this question A buffer solution contains 0.335 M NaHCO3 and 0.250 M K CO3. Determine the pH change when 0.097 mol NaOH is added to 1.00 L of the buffer pH after addition - pH before addition = pH change

If you need anymore values let me know! Thanks so much for your help!

Answer 1

From the Henderson- Hasselbalch equation for both acidic buffer and alkalined buffer,

$pH=pK_a+log\frac{[conjugate\, base]}{[acid]}$

$pOH=pK_b+log\frac{[conjugate\, acid]}{[base]}$

Case 1:

Here, acid is NaHCO₃ or HCO₃^- and conjugate base is K₂CO₃ or CO₃^2-. So, we have to use second pK_a of H₂CO₃. So now, pK_a = - log K_a2 = - log (4.8 x 10^-11) = 10.32

So,

$pH=10.32+log\frac{0.250}{0.335}=10.19$

Now, if 0.097 mol of NaOH is added in 1 L of solution, then it will neutralize 0.097 mol of acid. Then,

[CO₃^2-] = (0.250 + 0.097) M = 0.347 M

[HCO₃^-] = (0.335 - 0.097) M = 0.238 M

(M means mol.L^-1, since the solution volume is 1 L, we can easily add or subtract them)

Thus, new pH,

$pH=10.32+log\frac{0.347}{0.238}=10.48$

So, pH change is = (10.48 - 10.19) = 0.29

Case 2:

Here base = CH₃NH₂ and conjugate acid = CH₃NH₃Cl.

pK_b = - log K_b = - log (5 x 10^-4) = 3.30

$pOH=3.30+log\frac{0.449}{0.206}=3.64$

After addition of 0.047 mol of NaOH, it will neutralize that much mol of the conjugate acid.

Thus,

[CH₃NH₃Cl] = (0.449 - 0.047) M = 0.402 M

[CH₃NH₂] = (0.206 + 0.047) M = 0.253 M

Thus, new pH,

$pOH=3.30+log\frac{0.402}{0.253}=3.50$

Now, initial pH = 14 - initial pOH = 10.36

final pH = 14 - final pOH = 10.50

Thus, change in pH = 0.14