Question

A solution of pH 5 is diluted 100 times. Find the pH of the resulting solution.

Answer 1

@Vincent-Lyon.Fr
will it be 7 exactly, a little less or more? plz xplain

Answer 2

http://www.nvcc.edu/home/rgorham/sites/courses/101/101_powerpoint/7bio(polar_03_02_bw)/sld034.htm

Answer 3

No, pH = 7 would be for infinite dilution (nearing pure water).

The new pH value will depend on the nature of the acid involved.

For strong acids, the new pH will be 6.79

Answer 4

here's a good reference on pages 51 and on
http://finedrafts.com/files/campbellbio9th/CH2-3%20Chemistry%20and%20Water.pdf

Answer 5

plz explain a bit more
@Vincent-Lyon.Fr
why?

Answer 6

pH is a log scale (log is like, going up by a factor of 10 instead of a factor of 1) so if you were to dilute something to 1/10th it's original acidity, you would decrease the concentration of H+ ions by 10, but increase the pH by 1. Dilution by 10^2 (or 100) will decrease the concentration of H+ ions by 100, so pH increases by 2.

Answer 7

What lizcody1 says is true for strong acids, as long as you remain at least one step away from pH=7.
When you get between 6 and 7, water will level out the pH value, bringing in amounts of \(H_30^+\) ions of the same order of magnitude as those brought by the acid itself.

Answer 8

@shayanreloaded: here is the elaboration you are asking for.

A strong acid of pH 5 can be considered as 1 litre of pure water in which \(10^{-5}\) mole of a strong acid have been dissolved.

The pH 5 solution, diluted 100 times, is actually a \(10^{-7}\)M solution.

Water undergoes a self-ionisation reaction, resulting in an equilibrium in which:
\([H_3O^+]_{eq}.[OH^-]_{eq}=K_w=10^{-14}\) at 25°C.



Real question is then:
What is the pH of a \(c_o\) strong acid solution? (\(c_o=10^{-7}\)M in the end)

Here it goes:

Amounts in reaction are:

\(\large
\
\begin{array}{II}
& 2H_2O&=&H_3O^+&+&OH^- \\
n_i & &&c_o&&0 \\
\Delta n &&&+\epsilon &&+\epsilon\\
n_f & &&c_o+\epsilon &&\epsilon\\

\end{array}

\)

Let \(\large [H_3O^+]_{eq}=h=c_o+\epsilon\), then \(\large [OH^-]_{eq}=\epsilon=h-c_o\)

Equilibrium of water commands that:

\([\large H_3O^+]_{eq}.[OH^-]_{eq}=K_w\) , so

\(\large h(h-c_o)=K_w\) , which is a quadratic equation

\(\large h^2-c_oh-K_w=0\)

Positive solution is:

\(\large [H_3O^+]_{eq}=h=\Large\frac{c_o}{2}+\sqrt{\frac{c_o^2}{4}\large +K_w}\)

Plug in any value of \(c_o\) and you will get \(h\) and \(pH=-\log h\)

Note that if \(c_o>>\sqrt{K_w}=10^{-7}\) , result will give \(h\approx c_o\) , which is the usual result for a strong acid solution.

On the contrary, if, \(c_o<<\sqrt{K_w}=10^{-7}\) , equation will yield pH = 7.

Only the cases with \(c_o\approx 7\) are problematic and need the use of the quadratic equation above.

I will attach an Excel file with the graph \(pH = f\;(-\log c_o) \)

Answer 9

Thanks a lot