Question

The $ pH $ of milk of magnesia, a suspension of solid magnesium hydroxide in a saturated aqueous solution is $ 10.52 $ . What is the molarity of $ O{H^ - } $ ions in the saturated aqueous solution? The suspended undissolved $ Mg{(OH)_2} $ does not affect the measurement.
(A) $ 3 \times {10^{ - 4}}M $
(B) $ 1.7 \times {10^{ - 4}}M $
(C) $ 2.5 \times {10^{ - 4}}M $
(D) $ 3.6 \times {10^{ - 4}}M $

Answer 1

Hint: if the $ pH $ of the solution is given, $ pOH $ can be easily calculated using the formula:
 $ pH + pOH = 14 $
After getting the value of $ pOH $ , $ O{H^ - } $ ion can be calculated using the formula:
 $ pOH = - \log [O{H^ - }] $

Complete step by step solution:
In this question, we are given the $ pH $ of solution (milk of magnesia). To calculate the concentration of $ O{H^ - } $ ion, we need to find $ pOH $ by using the formula:
 $ pH + pOH = 14 $
Substituting the value of $ pH $ as given in the question,
 $ \Rightarrow 10.52 + pOH = 14 $
 $ \Rightarrow pOH = 14 - 10.52 $
 $ \Rightarrow pOH = 3.48 $
Now, we have the value of $ pOH $ . So, we can calculate the concentration of $ O{H^ - } $ ion by using the formula:
 $ pOH = - \log [O{H^ - }] $
On rearranging the formula, we get;
 $ \Rightarrow - pOH = \log [O{H^ - }] $
 $ \Rightarrow [O{H^ - }] = {10^{ - pOH}} $
Substituting the value of $ pOH $ ,
 $ \Rightarrow [O{H^ - }] = {10^{ - 3.48}} $
 $ \Rightarrow [O{H^ - }] = 0.0003 $
 $ \Rightarrow [O{H^ - }] = 3 \times {10^{ - 4}}M $
Hence, the molarity of $ O{H^ - } $ ion in the saturated aqueous solution is $ 3 \times {10^{ - 4}}M $ .
So, the correct option is option (A).

Additional information:
Magnesium is a naturally occurring mineral. Milk of magnesia is used as a laxative to relieve occasional constipation. It is also used as an antacid to relieve indigestion, sour stomach, and heartburn. Milk of magnesia works by absorbing water and neutralizing acid.

Note:
In any question, if we are given $ pH $ or $ pOH $ of solution then we can easily calculate the concentration of $ O{H^ - } $ ions in the solution by using the appropriate formula mentioned above.
Calculation mistakes should be avoided while taking antilogarithms in some cases.