Question

In: Chemistry

Consider the titration of a 26.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr....

Consider the titration of a 26.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Determine each of the following.

1. the pH at 4.0 mL of added acid

2. the pH at one-half of the equivalence point

3. the pH at the equivalence point

4. the pH after adding 5.0 mL of acid beyond the equivalence point

Solutions

Expert Solution

millimoles of CH3NH2 = 26.0 x 0.180 = 4.68

pKb of CH3NH2 = 3.38 standard value

1) millimoles of HBr added = 4.0 x 0.155 = 0.62

4.68 - 0.62 = 4.0 millimoles base left

0.62 millimoles salt formed

[base] = [4.0 / 30 ] = 0.13 M

[salt] = 0.62 / 30 = 0.021 M

pOH = pKb + log [salt] / [base]

pOH = 3.38 + log [0.021] / [0.13]

pOH = 2.60

pH = 14 - 2.60

pH = 11.40

2) at one-half of the equivalence point

pOH = pKb

pOH = 3.38

pH = 14 - 3.38

pH = 10.62

3) at equivalence point

all base becomes salt

4.68 millimoles HBr must be added

4.68 = V x 0.155

V = 30.1 mL HBr added

[salt] = 4.68 / 26 + 30.1 = 4.68 / 56.1 = 0.083 M

pOH = 1/2 [pKw + pKb + log C]

pOH = 1/2 [14 + 3.38 + log 0.083]

pOH = 8.15

pH = 14 - 8.15

pH = 5.85

4) millimoles of exess HBr = 5.0 x 0.155 = 0.775

total volume = 26 + 30.1 + 5.0 = 61.1 mL

.[HBr] = 0.775 / 61.1 = 0.013 M

as HBr strong base

[HBr] = [H+] = 0.013 M

pH = - log [H+]

pH = - log [0.013]

pH = 1.89


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