Question

Find the $$pH$$ of $$0.005M Ba{(OH)}_{2}$$ solution at $$25℃$$. Also calculate the $$pH$$ value of the solution when $$100ml$$ of the above solution is diluted to $$1000ml$$. (Assume complete ionization of barium hydroxide).

A
8,9
B
2,3
C
10,12
D
12,11

Solution

The correct option is D $$12,11$$              $$Ba{ \left( OH \right) }_{ 2 }\longrightarrow { Ba }^{ 2+ }+2{ OH }^{ - }$$Initial$$-0.005M$$                $$0$$              $$0$$               $$0$$                   $$0.005M$$      $$2\times0.005M$$$$\therefore \quad \left[ { OH }^{ - } \right] =0.01M$$$$\therefore \quad pOH=-\log { \left[ { OH }^{ - } \right] } =-\log { \left( 0.01 \right) } =2$$But $$pH+pOH=14$$$$\therefore pH=12$$$$\therefore \quad { \left[ { H }^{ + } \right] }_{ 1 }={ 10 }^{ -pH }={ 10 }^{ -12 }M$$Now, $${ M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 }$$$${ 10 }^{ -12 }\times 100={ M }_{ 2 }\times 1000$$$$\therefore \quad { M }_{ 2 }={ 10 }^{ -11 }M$$$$\therefore \quad { \left[ { H }^{ + } \right] }_{ 2 }={ 10 }^{ -11 }M$$$$pH$$ of solution when diluted $$=-\log { \left[ { 10 }^{ -11 } \right] }$$                                                 $$=11$$Chemistry

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