Solveeit Logo
Login

Question

Question: Calculate the pH of a solution made by mixing \[50\; mL\;{\rm{ }} \;of \;{\rm{ }}0.01M\]\( Ba{\left(...

Calculate the pH of a solution made by mixing 50;mL;;of;0.01M50; mL;{\rm{ }} ;of ;{\rm{ }}0.01MBa(OH)2 Ba{\left( {OH} \right)_2} with 50 mL water.

Explanation

Solution

First we have to calculate the total number of moles. After moles calculation we have to calculate the concentration which provides us the concentration of OH−O{H^ - } ions.

Complete step by step answer:
We know that every solution whether acidic, alkaline, or neutral, contains both H+{H^ + } and OH−OH^- ions. The product of their concentration’s ions is always constant equal to 1×10−141 \times {10^{ - 14}} at 250C{25^0}C. Whether the solution is acidic or alkaline it depends upon which of the two ions is present in greater concentration than the other. But, since knowing the concentration of one of these ions, that of the other can be calculated, it is convenient to express acidity or alkalinity of a solution by referring to the concentration of hydrogen ions only. Now H+{H^ + } ion concentration can vary within wide limits, usually from 1 mole per litre (as in 1M HCl) to about 10−14{10^{ - 14}} mole per litre (as in NaOHNaOH).
pH is defined as the logarithm (to the base 10) of the concentration (in moles per litre) of hydrogen ion.
pH=−log⁡[H+]=−log⁡[H3O+]pH = - \log \left[ {{H^ + }} \right] = - \log \left[ {{H_3}{O^ + }} \right]
In the question we have 50 mL of 0.01M Ba(OH)2Ba{\left( {OH} \right)_2} which is mixed with 50 mL water.
So, moles of Ba(OH)2=50×0.01=0.5  molesBa{\left( {OH} \right)_2} = 50 \times 0.01 = 0.5\;{\rm{moles}}.
Since the solution is mixed with water i.e. it is diluted with 50 mL water, thus total volume becomes 100 mL.
Concentration=Number  of  molesTotal  Volume{\rm{Concentration}} = \dfrac{{{\rm{Number}}\;{\rm{of}}\;{\rm{moles}}}}{{{\rm{Total}}\;{\rm{Volume}}}}
Concentration=0.5100=0.005{\rm{Concentration}} = \dfrac{0.5}{100} = 0.005
So, the concentration of Ba(OH)2Ba{\left( {OH} \right)_2} ionises into ;
                          Ba(OH)2→Ba2+ + 2OH− Initial conc;    0.005                0              0 Final conc;      0                       0.005     2×0.005 [OH]−=0.010=1×10−2 pOH = −log  (1×10−2) ∴pOH=2 ∴pH=14−2=12\begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Ba{\left( {OH} \right)_2} \to B{a^{2 + }}\, + \,2O{H^ - }\\\ Initial\,conc;\;\;0.005\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\\ Final\,conc;\;\;\;0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.005\,\,\,\,\,2 \times 0.005\\\ {\left[ {OH} \right]^ - } = 0.010 = 1 \times {10^{ - 2}}\\\ pOH\,=\, - log\;(1 \times {10^{ - 2}})\\\ \therefore pOH = 2\\\ \therefore pH = 14 - 2 = 12 \end{array}
Hence the pH of a solution made by mixing 50 mL of 0.01M Ba(OH)2Ba{\left( {OH} \right)_2} with 50 mL water is 12.

Note: pH of neutral solution is 7. pH of acidic solutions is less than 7 and pH of basic solutions is greater than 7. pOH of solution is calculated by simply subtracting it from 14.