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October 20, 2014

October 20, 2014

Posted by **AAA** on Monday, August 6, 2012 at 7:49am.

- Chemistry -
**DrRuss**, Monday, August 6, 2012 at 8:02amI 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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