1. Equal molar amounts of nitric acid (a strong acid) and methylamine (a weak ba

Question

1. Equal molar amounts of nitric acid (a strong acid) and methylamine (a weak base) are combined in an aqueous solution. The pH of the resulting solution is: (A. greater than 7 B. close to 7 OR C. less than 7)

2. A 1.00 L aqueous solution contains 0.10 mol acetic acid and 0.10 mol sodium acetate. The initial pH is 4.74. Calculate the pH after the addition of 0.051 mol of OH-. (Watch for sig figs please)

3. Using the Henderson-Hasselbalch equation, calculate the pH of a buffer solution that contains 0.62 M NH4Cl and 0.36 M NH3. (Watch for sig figs please)

Explanation / Answer

1)

Since this is mixture of equimolar strong acid and weak base, solution will be acidic

This is because HNO3 and CH3NH2 reacts as:

HNO3 + CH3NH2 —> CH3NH3+ + NO3-

So, At equilibrium, there is weak acid CH3NH3+ present and hence solution is acidic

pH will be less than 7

Answer: C

2)

When there are equal mol of CH3COOH and CH3COONa

pH = pKa

So, pKa = 4.74

mol of OH- added = 0.051 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.1 M *1.0 L

mol of CH3COO- = 0.1 mol

mol of CH3COOH = 0.1 M *1.0 L

mol of CH3COOH = 0.1 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.1 + 0.051) mol

mol of CH3COO- = 0.151 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.1 - 0.051) mol

mol of CH3COOH = 0.049 mol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.74+ log {0.151/4.9*10^-2}

= 5.229

Answer: 5.23

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