A solution has a pOH of 6.39. What is the solutions pH? (You must answer to the correct number of s.d.'s)

For the acid-base equilibrium that operates in water.....

#2H_2O rightleftharpoons H_3O^+ + HO^-#

We know that at under standard conditions, #298*K, "1 atmosphere"#.............

#K_w=[H_3O^+][HO^-]=10^-14#

This defines the acid-base equilibrium, the autoprotolysis of water.

We can take #log_10# of BOTH sides........

#log_10K_w=log_10[H_3O^+]+log_10[HO^-]#

And on rearrangement,

#underbrace(-log_10[H_3O^+])_(pH)-underbrace(log_10[HO^-])_(pOH)=underbrace(-log_10K_w)_(pK_w)#

Given that #K_w=10^-14#, and #pH=-log_10[H_3O^+]#, then BY DEFINITION, #underbrace(-log_10K_w)_(pK_w)=-log_(10)10^(-14)=-(-14)=14#, and the defining relationship, which you may not have to derive, but WILL have to remember,

#14=pH +pOH#

And given that #pOH=6.39#, this means that #[HO^-]=10^(-6.39)*mol*L^-1#, and #pH=14-6.39=7.61#, and (FINALLY) #[H_3O^+]=10^(-7.61)*mol*L^-1=2.4xx10^-8*mol*L^-1.#

Now this might seem a lot of work, but only because I derived the equation. You must be able to use the relationship...........

#14=pH +pOH#

These logarithmic terms were introduced back in the day, before the advent of electronic calculators. Log tables, printed values of #log_10# and #log_e# were widely used by scientists, engineers, and by students of course.

For this question, you can use the rule:
#pH +pOH=14#

Rewritten this gives us:
#pH=14-pOH#
#pH=14-6.39#
#pH=7.61#

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How is #pH+pOH =14# established?

In water, the following (ionization) reaction occurs:
#2 H_"2"O -> H_"3"O^"+" + OH^"-"#

Therefore the equilibrium can be written like
#K_"c"=([H_"3"O^"+"]*[OH^"-"])/[H_"2"O]#
Since water is the solvent here, we do not take the concentration of water into consideration (is very big). We obtain the following expression:
#K_"c"=[H_"3"O^"+"]*[OH^"-"]#

The #K_"c"# in this equation represents a special number because we talk about the ionisation of water. Therefore we denote #K_"c"# as #K_"w"#. The value of the #K_"w"# is measured at 25°C.
#K_"w" (25°C) = 1*10^(-14)#
This means we can say:
#K_"c"=K_"w"=[H_"3"O^"+"]*[OH^"-"]=1*10^(-14)#

To get from the #[H_"3"O^"+"]# (concentration #H_"3"O^"+"#) to the pH, we use the following formula:
#pH= - log[H_"3"O^"+"]#
The same is true for the #[OH^"-"]#, since we define pOH as
#pOH=- log[OH^"-"]#

Now if we take the Log from both sides of the #K_"w"# equation, we get:
#log(1*10^(-14))=log([H_"3"O"]*[OH^-])#
A mathematics rule tells us that multiplying inside the logarithm function is the same as adding these logarithms. Therefore we get
#log(10^(-14))=log[H_"3"O"]+log[OH^-]#

And now we can use the definitions of pOH and OH! We get:
#log(10^(-14))=-pH - pOH#
with #log(10^(-14))=-14# we get the function
#-pH-pOH =-14#
Which is the same as
#pH+pOH=14#