CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
2,3
loader
C
10,12
loader
D
12,11
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image