Page 179 - Modern Analytical Chemistry

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162 Modern Analytical Chemistry

–
Since the [F ] is 2.6% of C HF , this assumption is also within our limit that the error
+
be no more than ±5%. Accepting our solution for the concentration of H 3 O , we
find that the pH of 1.0 M HF is 1.59.
How does the result of this calculation change if we require our assumptions to
have an error of less than ±1%. In this case we can no longer assume that [HF] >>
–
[F ]. Solving the mass balance equation (6.37) for [HF]
–
[HF] = C HF –[F ]
and substituting into the K a expression along with equation 6.39 gives
+ 2
[ HO ]
3
K a = +
C HF –[ HO ]
3
Rearranging leaves us with a quadratic equation
+ 2
+
[H 3 O ] = K a C HF – K a [H 3 O ]
+ 2
+
[H 3 O ] + K a [H 3 O ]– K a C HF =0
which we solve using the quadratic formula
2
– b ± b – 4 ac
x =
a 2
2
where a, b, and c are the coefficients in the quadratic equation ax + bx + c =0.
Solving the quadratic formula gives two roots, only one of which has any chemical
significance. For our problem the quadratic formula gives roots of

10
) – ( )( )(– .68
– .68 ´10 –4 ± ( .68 ´10 –4 2 4 1 ´0 –4 )( . )
1
x =
21
()
–.68 ´10 –4 ± .522 ´10 –2
=
2
2
= .57 ´10 –2 or – .63 ´10 –2
2
Only the positive root has any chemical significance since the negative root implies
+
+
that the concentration of H 3 O is negative. Thus, the [H 3 O ] is 2.6 ´10 –2 M, and
the pH to two significant figures is still 1.59.
This same approach can be extended to find the pH of a monoprotic weak
base, replacing K a with K b , C HF with the weak base’s concentration, and solving for
–
+
the [OH ] in place of [H 3 O ].
EXAMPLE 6.11
Calculate the pH of 0.050 M NH 3 . State any assumptions made in simplifying
the calculation, and verify that the error is less than 5%.
SOLUTION
–5
Since NH 3 is a weak base (K b = 1.75 ´10 ), we assume that
–
+
[OH ] >> [H 3 O ] and C NH 3 = 0.050 M
With these assumptions, we find (be sure to check the derivation)
[OH – ] = KC = ( .175 ´10 –5 )( .050 ) =935 ´ 10 –4 M
0
.
bNH 3