Question

Calculate the pH of the resulting solution if

21.0 mL of 0.210 M HCl(aq) is added to 26.0mL of

0.210M NaOH(aq) .

pH=

Calculate the pH of the resulting solution if 21.0mL of

0.210 M HCl(aq) is added to 31.0mL of 0.260 M NaOH(aq).

pH=

Answer 1

The balanced neutralisation reaction is,

HCl (aq) + NaOH (aq.) -----------> NaCl (aq.) + H2O (l)

In both the cases, MBVB > MAVA

1.

MA = 0.210 M

VA = 21.0 mL

MB = 0.210 M

VB = 26.0 mL

Formula is,

[OH-] =

[OH-] = [(0.210 * 26.0) - (0.210 * 21.0)] / (21.0 + 26.0)

[OH-] = 0.0223 M

pOH = - Log[OH-] = - Log(0.0223) = 1.65

Therefore, pH + pOH = 14

pH = 14 - 1.65

pH = 12.35

2.

MA = 0.210 M

VA = 21.0 mL

MB = 0.210 M

VB = 31.0 mL

Formula,

[OH-] =

[OH-] = [(0.210 * 31.0) - (0.210 * 21.0)] / (21.0 + 31.0)

[OH-] = 0.0404 M

pOH = - Log[OH-] = - Log(0.0404) = 1.39

Therefore, pH + pOH = 14

pH = 14 - 1.39

pH = 12.61