A couple of things about this last step: First of all, the reaction indicates that for
every 1 HF molecule that dissociates, 1 H+ and 1 F− are produced. That’s why
we used just x’s in the algebraic expression instead of using three unrelated
variables. Second of all, the equation above is actually a quadratic equation and
can therefore be solved using the quadratic equation. However, in cases in which
[HA] is at least 3 orders of magnitude (10^3 ) larger than Ka, the fraction of HA
lost due to dissociation is tiny and is completely ignored by chemists (this same
rule applies for weak bases). So, for this problem, we can now write
(7 × 10−4) =
Fourth, solve the algebraic expression.(7 × 10−4) =
x^2 = (7 × 10−4) × (1.5 M) = 1 × 10−3x = 1 × 10−1.5Fifth, don’t worry about the fractional exponent
because it will disappear when you calculate pH.pH = −log [H+]
= −log (1 × 10−1.5 M)
= 1.5Try another example to make sure you’ve got it.
Example: Given that Kb(NH3) = 1.8 × 10−5, what is the pH of 0.5 M NH 3 (aq)?
Solution: First, write the balanced chemical equation.