Posted by AAA on Monday, August 6, 2012 at 7:49am.
I have a 40 ml solution of pH 4.36. I need to make this solution at pH 8.5 by adding extra 10 ml of HCl and NaOH.
Not sure that this question entirley makes sense. However,
if pH is 4.36 then pOH is 9.74 and [OH-] is 1.82 x 10^-10 M
if target pOH is 5.5 then [OH-] is 3.16 x 10^-6
so total number of milli moles of OH- in target solution is
50 x 3.16 x 10^-6 mmoles
and we have
40 x 1.82 x 10^-10 mmoles
so we need the difference which is
(50 x 3.16 x 10^-6 mmoles)-(40 x 1.82 x 10^-10 mmoles)
which is ca 50 x 3.16 x 10^-6 mmoles
=0.000158 mmoles in 10 ml so the concentration is
1.58 x 10-5 M of NaOH solution.
Please check the maths!
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