CHEM 102 Class 8
pH Calculations: Strong Acids and Bases
(1)* Calculate the pH of (a) 0.055 M HNO3 (b) 4.55 M HCl
Both are strong acids
(a) HNO3 is a strong acid HNO3(aq) → H+(aq) + NO3-(aq). Therefore the concentration
of H+ in the final solution is 0.055 M
pH = -log[H+] = -log(0.055) = 1.26
(b) HCl(aq) → H+(aq) + Cl-(aq). Therefore the concentration of H+ in the final solution
is 4.55 M
pH = -log[H+] = -log(4.55) = -0.66
(2)* Calculate the pH of 0.10 M NaOH
NaOH is a strong base so we should work in terms of pOH NaOH(aq) → Na+(aq) + OH(aq). Therefore the concentration of OH- in the final solution is 0.10 M.
pOH = -log[OH-] = -log(0.10) = 1.00
pKW = pH + pOH = 14.00
pH = 14.00 – 1.00 = 13.00
(3)** A solution has a pH of 6.88. What is [H3O+]? [OH-]?
pH = − log(H+ )
[H+ ] = 10−pH = 10−6.88 = 1.32x10−7
[OH− ]⋅ [H+ ] = K w = 1.00x10−14
[OH− ] =
Kw
+
[H ]
=
1.00x10−14
−7
1.32x10
= 7.59x10−8
(4)** Very concentrated nitric acid has a pH of -1.2. What is [H3O+]?
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pH = − log(H+ )
[H+ ] = 10−pH = 10−(−1.2) = 15.8 M
(5)** A solution has an [OH-] of 1.4x10-7 M. Show that this solution is slightly basic by
calculating its pH.
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pOH = − log(OH− ) = − log(1.4x10−7 ) = 6.85
pK w = pH + pOH = 14.0
pH = 14.0 − pOH = 7.15
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Since the pH is greater than 7.00 (neutral) the concentration of hydroxide exceeds
hydronium and the solution is basic
(6)** Like any equilibrium constant, Kw varies with temperature. Its value at 37 °C is
2.4x10-14. What is the pH of (neutral) pure water at 37 °C?
K w = [H3O+ ]⋅ [OH− ] = 2.4x10−14
[H3O+ ]⋅ [H3O+ ] = 2.4x10−14
[H3O+ ] = 2.4x10−14 = 1.55x10−7
pH = − log(1.55x10−7 ) = 6.81
(7)** A solution has a [H+] of 6.7x10-4 M. What are [OH-] and pOH?
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pH = − log(H+ ) = − log(6.7x10−4 ) = 3.17
pOH = pK w − pH = 14.00 − 3.17 = 10.83
[OH− ] = 10−pOH = 10−10.83 = 1.48x10−11
pH Calculations: Weak Acids and Bases
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(8)** Calculate (a) the pH and (b) percent ionization of 1.0 M acetic acid if the Ka is
1.86x10-5 at 20 °C.
(a) This is a weak acid CH3CO2H _ CH3CO2- + H+. Writing an expression for Ka gives
−
[H+ ]⋅ [CH3CO2 ]
Ka =
= 1.86x10−5
[CH3CO2H]
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We’ll assume that H3O+ comes only from the acetic acid (not from water) and that the
dissociation is small so the concentration of the acetic acid is unchanged, both reasonable
given the size of Ka.
−5
1.86x10
[H+ ]⋅ [CH3CO2 ] [H+ ]⋅ [H+ ] [H+ ]2
=
=
=
[CH3CO2H]
[CH3CO2H]
1.0
−
[H+ ]2 = 1.86x10−5 ×1.0
[H+ ] = 1.86x10−5 = 4.31x10−3
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Checking the assumptions:
H+ from water (10-7 M / 4.31x10-3 M) x 100 = 0.0023%
Initial [CH3CO2H] unchanged (4.31x10-3 M / 1.0 M) x 100 = 0.43 %
The last step is to calculate pH
pH = − log(H+ ) = − log(4.31x10−3 ) = 2.37
(b) The percent ionization is defined as
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% ionization =
[H + ]eqm
[HA]initial
4.31x10−3 M
× 100 =
× 100 = 0.43%
1.0 M
(9)** Calculate the pH of 0.004 M solution of vitamin C (ascorbic acid, Ka is 1.0x10-5 at
20 °C).
This is a weak acid AscH _ Asc- + H+. Writing an expression for Ka gives
Ka =
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[H+ ]⋅ [Asc− ]
= 1.0x10−5
[AscH]
We’ll assume that H3O+ comes only from the ascorbic acid (not from water) and that the
dissociation is small so the concentration of the acetic acid is unchanged, both reasonable
given the size of Ka.
−5
1.0x10
[H+ ]⋅ [Asc− ] [H+ ]⋅ [H+ ] [H+ ]2
=
=
=
[AscH]
[AscH]
0.004
[H+ ]2 = 1.0x10−5 × 0.004
[H+ ] = 4.0x10−8 = 2.0x10−4
Checking the assumptions:
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H+ from water (10-7 M / 2.0x10-4 M) x 100 = 0.05%
Initial [AscH] unchanged (2.0x10-4 M / 0.004 M) x 100 = 5.0 % Note that this is the
limit of the error we can tolerate in the second case. If the concentration was any smaller,
we’d have to solve a quadratic.
The last step is to calculate pH
pH = − log(H+ ) = − log(2.0x10−4 ) = 3.7
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(10)*** Calculate the [H+] of 0.20 M phosphoric acid (H3PO4) if the Ka1 is 7.2x10-3, Ka2 =
6.3x10-8, Ka3 = 4.2x10-13 at 25 °C.
This is a weak polyprotic acid that only the first ionization, Ka1, effectively contributes to
the [H+] H3PO4 _ H2PO4- + H+. Writing an expression for Ka gives
−
[H+ ]⋅ [H2PO4 ]
Ka 1 =
= 7.2x10−3
[H3PO4 ]
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We’ll assume that H3O+ comes only from the phosphoric acid (not from water) and that
the dissociation is small so the concentration of the phosphoric acid is unchanged, both
reasonable given the size of Ka.
−3
7.2x10
[H+ ]⋅ [H2PO4 ] [H+ ]⋅ [H+ ] [H+ ]2
=
=
=
[H3PO 4 ]
[H3PO4 ]
0.2
−
[H+ ]2 = 7.2x10−3 × 0.2
[H+ ] = 1.44x10−3 = 3.79x10−2
Checking the assumptions:
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H+ from water (10-7 M / 3.79x10-2 M) x 100 = 0.00026%
Initial [H3PO4] unchanged (3.79x10-2 M / 0.2 M) x 100 = 18.95 %
The last step shows that the second assumption is not valid. Hence we must rewrite the
equation for Ka1 as
−3
7.2x10
[H+ ]⋅ [H2PO 4 ]
[H+ ]⋅ [H+ ]
[H+ ]2
=
=
=
[H3PO4 ]
[H3PO4 ]init − [H+ ] 0.2 − [H+ ]
−


7.2x10−30.2 − [H+ ] = [H+ ]2


0.00144 − 7.2x10−3[H+ ] = [H+ ]2
0 = [H+ ]2 + 7.2x10−3[H+ ] − 0.00144
This is a quadratic equation with solutions 0.0345 and -0.0417. Clearly the negative root
is inappropriate.
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Note the difference in [H+] is significant at ((0.0379 – 0.0345) / 0.0345) x 100 = 9.9%