Q4. 1.0 mL of 1.0 M NaOH is added to 100 mL of the buffer of Example 3 (0.10 M acetic acid, 0.10 M sodium acetate). Find the new pH and compare it with adding the same NaOH to 100 mL of water.
The OH− (0.0010 mol) turns 0.0010 mol of acetic acid into acetate: 0.0090 mol acid and 0.011 mol acetate, so pH = 4.76 + log(0.011/0.0090) ≈ 4.85, a rise of only 0.09. In water, [OH−] = 0.0010/0.101 = 9.9 × 10−3 M and the pH rises from 7.00 to about 12.00.
Q5. Which of these, each in 1 L, is a buffer: (a) 0.10 mol HCl + 0.10 mol NaCl; (b) 0.20 mol NH3 + 0.10 mol HCl; (c) 0.10 mol NH3 + 0.10 mol HCl? Find the buffer’s pH (pKb of NH3 = 4.75).
Only (b): the HCl turns 0.10 mol of NH3 into NH4+, leaving 0.10 mol of each, so pH = 9.25. (a) has no weak partner; (c) is NH4Cl alone.
Q6. What mass of sodium acetate (82 g mol−1) must be dissolved in 1.0 L of 0.10 M acetic acid to make a buffer of pH 4.50 (pKa = 4.76)? Ignore the change in volume.
[salt]/[acid] = 104.50 − 4.76 = 10−0.26 ≈ 0.55, so [salt] = 0.055 M: 0.055 mol × 82 g mol−1 ≈ 4.5 g; less salt than acid because pH < pKa.
Q7. A buffer is 0.20 M in NH3 and 0.10 M in NH4Cl (pKb = 4.75). Find its pH, and say whether adding more NH4Cl raises or lowers it.
pOH = 4.75 + log(0.10/0.20) = 4.45, so pH = 9.55. Adding NH4Cl increases [BH+], raises pOH and so lowers the pH.
Q8. A student uses NCERT’s Eq. 6.42 as printed, pH = pKa + log([BH+]/[B]), for the buffer in Q7. What does she get, and how could she spot the error?
9.25 + log(0.10/0.20) = 8.95, the wrong value. A buffer with more base than conjugate acid must have pH above pKa (9.25), so 8.95 cannot be right; the correct equation, with [B]/[BH+], gives 9.55.