What is the pH of a buffer made from 0.210 mol of HCNO (Ka = 3.5 × 10⁻⁴) and 0.410 mol of NaCNO in 2.0 L of solution?

Answer 1

pKa = -log Ka

pKa = -log 3.5x10-4

pKa = 3.46

Use the Henderson Hasselbalch equation

pH = pKa + log [NaCNO]/[HCNO]

[NaCNO] = 0.410 mol/2 L = 0.205 M

[HCNO] = 0.210 mol/2 L = 0.105 M

pH = 3.46 + log (0.205/0.105) = 3.46 + 0.29

pH = 3.75