Can anyone get this? Sorry I can't post a picture of it.
Basically I am confused with how if 2 concentrations of an acid at 2 different points differ by a factor of 2, it means the pH of the 2 solutions differs by -log sqrt2
HOW?!
From the shape of the curve we know that the acid involved is a weak acid. Because a weak acid does not fully dissociate, we cannot simply use pH = -log[HA], because [HA]
added does not equal [H
+]. You need to use the BR weak acid equation. The concentration of the weak acid at the start of titration (point a) is double that of the weak acid formed at equivalence in the titration of an equimolar quantity of its weak base (point d). So we need only include a factor of 2 in the concentration term of the equation.
pH = 1/2pKa - 1/2log[HA]
The pK
a doesn't change, so only the log[HA] term needs to be considered. The difference is in the
1/
2log[HA] term, which results in a difference of
1/
2log(2) which is the log of square root of 2.
Cocneptually you should think that doubling the concentration of a strong acid would lower the pH by log2, but with a weak acid there is less dissociation, so the pH would not go down by as much. The math gives us the exact number as being log(root 2).
p simply means "negative log" so pH means -log[H+]. Or in mathematical terms: pH=-log[H+].
For example, increasing the H+ concentration by a factor 10 will lower the pH by 1. This is because of the logarithm that has a base ten.
So it WILL NOT differ by -log sqrt2, it will differ by -log 2. This is because if we plug a 2 into the equation, pH=-log[2H+] or in expanded form pH=-log[H+] -log(2).
Hope this helps.
You make a valid point if it were a question about two solutions of hydronium (or two strong acid solutions). But the question involves a weak acid and not a strong acid or hydronium solutions, which makes it a little more challenging than the strong acid case.
