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Question: Find the hydroxide concentration for a solution with pH of \(10\) at \({{25}^{0}}C\) A. \({{10}^{...

Find the hydroxide concentration for a solution with pH of 1010 at 250C{{25}^{0}}C
A. 10−4M{{10}^{-4}}M
B. 10−10M{{10}^{-10}}M
C. 10−4M{{10}^{-4}}M
D. 10−1M{{10}^{-1}}M

Explanation

Solution

Hydroxide concentration means the number of or the amount of hydroxide ion that is OH−O{{H}^{-}} in a solution.

Complete step by step answer:
As we know, the pH can be calculated as the negative of log of hydrogen ion concentration that is pH=−log⁡10[H+(aq)]pH=-{{\log }_{10}}[{{H}^{+}}(aq)].
Similarly, pOH is negative of log of hydroxide ion concentration in aqueous medium of course, which means pOH=−log⁡10[OH−(aq)]pOH=-{{\log }_{10}}[O{{H}^{-}}(aq)].
By taking the antilog of the above equations we can calculate the concentrations of hydrogen or hydroxide ions or we are simply going backwards to obtain the ionic concentrations. The concentration can only be found in aqueous medium since ions can only be separated in aqueous medium and not in solid or other mediums.
Here, we have to log to the base 1010 but if in question it is given in any other base we can still calculate it by using different log formulas but since nothing is mentioned in the question that means we can normally use this formula.
We also know that pH+pOH=14pH+pOH=14
Now,
pOH=−log⁡10[OH−(aq)] [OH−(aq)]=10−pOH [OH−(aq)]=10−(14−pH) [OH−(aq)]=10−(14−10) [OH−(aq)]=10−4 \begin{aligned} & pOH=-{{\log }_{10}}[O{{H}^{-}}(aq)] \\\ & [O{{H}^{-}}(aq)]={{10}^{-pOH}} \\\ & [O{{H}^{-}}(aq)]={{10}^{-(14-pH)}} \\\ & [O{{H}^{-}}(aq)]={{10}^{-(14-10)}} \\\ & [O{{H}^{-}}(aq)]={{10}^{-4}} \\\ \end{aligned}( from the above equation)

Therefore, the concentration of hydroxide ions is 10−4M{{10}^{-4}}M and hence option c is correct.

Note: There are other ways to solve this problem but this is the simplest since it doesn’t involve much of calculation and chances of mistakes are very rare.
You can first calculate the pOH and put it in the equation directly rather than replacing it in the equation but it will increase two steps and a further bothersome situation will arise.