Consider a solution of the acid corresponding to the last digit of your Student ID Number. The initial concentration of the acid is equal to 0.0045 M. Find the pH. Use the equilibrium constants from the Table of Acid Ionization Constants. Write the equilibrium reaction that your answer is based on and show all of your work.
0) Bromoacetic acid
1) Chloroacetic acid
2) Chlorous acid
3) Cyanic acid
4) Dichloroacetic acid
5) Fluoroacetic acid
6) Hydrogen sulfate ion
7) Hydrogen selenate ion
8) Iodic acid
9) Trichloroacetic acid
0) Bromoacetic acid
CH2BrCOOH(aq) + H2O(l) → ← H3O+(aq) + CH2BrCOO–(aq)
Ka = [H3O+]e [CH2BrCOO–]e [CH2BrCOOH]e = 1.3×10–3
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/1.3×10–3 = 0.35: No
1.3×10–3 = x2/(0.0045 – x)
x2 + 1.3×10–3x – 5.85×10–6 = 0
Using the quadratic equation gives x = 1.9×10–3 or –3.2×10–3
Only the positive value makes sense so [H3O+] = x = 1.9×10–3 M
pH = –log(1.9×10–3) = 2.72
1) Chloroacetic acid
CH2ClCOOH(aq) + H2O(l) → ← H3O+(aq) + CH2ClCOO–(aq)
Ka = [H3O+]e [CH2ClCOO–]e [CH2ClCOOH]e = 1.4×10–3
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/1.4×10–3 = 0.32: No
1.4×10–3 = x2/(0.0045 – x)
x2 + 1.4×10–3x – 6.3×10–6 = 0
Using the quadratic equation gives x = 1.9×10–3 or –3.3×10–3
Only the positive value makes sense so [H3O+] = x = 1.9×10–3 M
pH = –log(1.9×10–3) = 2.72
2) Chlorous acid
HClO2(aq) + H2O(l) → ← H3O+(aq) + ClO2–(aq)
Ka = [H3O+]e [ClO2–]e [HClO2]e = 1.1×10–2
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/1.1×10–2 = 0.41: No
1.1×10–2 = x2/(0.0045 – x)
x2 + 1.1×10–2x – 4.95×10–5 = 0
Using the quadratic equation gives x = 3.4×10–3 or –1.4×10–2
Only the positive value makes sense so [H3O+] = x = 3.4×10–3 M
pH = –log(3.4×10–3) = 2.47
3) Cyanic acid
HOCN(aq) + H2O(l) → ← H3O+(aq) + OCN–(aq)
Ka = [H3O+]e [OCN–]e [HOCN]e = 3.5×10–4
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/3.5×10–4 = 13: No
3.5×10–4 = x2/(0.0045 – x)
x2 + 3.5×10–4x – 1.575×10–6 = 0
Using the quadratic equation gives x = 1.1×10–3 or –1.4×10–3
Only the positive value makes sense so [H3O+] = x = 1.1×10–3 M
pH = –log(1.1×10–3) = 2.96
4) Dichloroacetic acid
CCl2HCOOH(aq) + H2O(l) → ← H3O+(aq) + CCl2HCOO–(aq)
Ka = [H3O+]e [CCl2HCOO–]e [CCl2HCOOH]e = 5.5×10–2
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/5.5×10–2 = 0.1: No
5.5×10–2 = x2/(0.0045 – x)
x2 + 5.5×10–2x – 2.475×10–4 = 0
Using the quadratic equation gives x = 4.2×10–3 or –5.9×10–2
Only the positive value makes sense so [H3O+] = x = 4.2×10–3 M
pH = –log(4.2×10–3) = 2.38
5) Fluoroacetic acid
CH2FCOOH(aq) + H2O(l) → ← H3O+(aq) + CH2FCOO–(aq)
Ka = [H3O+]e [CH2FCOO–]e [CH2FCOOH]e = 2.6×10–3
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/2.6×10–3 = 1.7: No
2.6×10–3 = x2/(0.0045 – x)
x2 + 2.6×10–3x – 1.17×10–5 = 0
Using the quadratic equation gives x = 2.4×10–3 or –5.0×10–3
Only the positive value makes sense so [H3O+] = x = 2.4×10–3 M
pH = –log(2.4×10–3) = 2.62
6) Hydrogen sulfate acid
HSO4–(aq) + H2O(l) → ← H3O+(aq) + SO42–(aq)
Ka = [H3O+]e [SO42–]e [HSO4–]e = 1.1×10–2
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/1.1×10–2 = 0.41: No
1.1×10–2 = x2/(0.0045 – x)
x2 + 1.1×10–2x – 4.95×10–5 = 0
Using the quadratic equation gives x = 3.4×10–3 or –1.4×10–2
Only the positive value makes sense so [H3O+] = x = 3.4×10–3 M
pH = –log(3.4×10–3) = 2.47
7) Hydrogen selenate acid
HSeO4–(aq) + H2O(l) → ← H3O+(aq) + SeO42–(aq)
Ka = [H3O+]e [SeO42–]e [HSeO4–]e = 2.2×10–2
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/2.2×10–2 = 0.2: No
2.2×10–2 = x2/(0.0045 – x)
x2 + 2.2×10–2x – 9.9×10–5 = 0
Using the quadratic equation gives x = 3.8×10–3 or –2.6×10–2
Only the positive value makes sense so [H3O+] = x = 3.8×10–3 M
pH = –log(3.8×10–3) = 2.42
8) Iodic acid
HIO3(aq) + H2O(l) → ← H3O+(aq) + IO3–(aq)
Ka = [H3O+]e [IO3–]e [HIO3]e = 1.6×10–1
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/1.6×10–1 = 0.03: No
1.6×10–1 = x2/(0.0045 – x)
x2 + 1.6×10–1x – 7.2×10–4 = 0
Using the quadratic equation gives x = 4.4×10–3 or –1.6×10–1
Only the positive value makes sense so [H3O+] = x = 4.4×10–3 M
pH = –log(4.4×10–3) = 2.36
9) Trichloroacetic acid
CCl3COOH(aq) + H2O(l) → ← H3O+(aq) + CCl3COO–(aq)
Ka = [H3O+]e [CCl3COO–]e [CCl3COOH]e = 3.0×10–1
Initial0.0045 0 0
Change– x +x +x
Equilibrium0.0045 – x x x
Approximate? (0.0045)/3.0×10–1 = 0.015: No
3.0×10–1 = x2/(0.0045 – x)
x2 + 3.0×10–1x – 1.35×10–3 = 0
Using the quadratic equation gives x = 4.4×10–3 or –3.0×10–1
Only the positive value makes sense so [H3O+] = x = 4.4×10–3 M
pH = –log(4.4×10–3) = 2.36