How do you determine the pH of a solution that is 3.85% KOH by mass? Assume that the solution has density of 1.01 g/mL.

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Answer 1 · 2016-12-16T19:07:38

#pH=-log_10[H_3O^+]=13.84#

In #1# #mL# of solution, #[NaOH]="moles of NaOH"/"volume of solution"#

#=(1.01*gxx3.85%)/(56.11*g*mol^-1)xx1/(1xx10^-3L)#

#=0.693*mol*L^-1#

Now #pOH=-log_10[HO^-]#, #pOH=-log_10(0.693)#

#=0.159#

But in water, we know that #pH+pOH=14#,

And thus #14-0.159=pH=13.84#

The high value of #pH# is reasonable for a solution of potash.

See here for more on the relationship between #pH# and acidity.