Question

$ 7.3grams $ of $ Hcl $ is dissolved in $ 20litre $ solution. $ 50ml $ of this solution is taken in $ 250ml $ flask and water is added up to the mark. $ 40ml $ if this diluted $ Hcl $ solution exactly neutralizes $ 20ml $ of $ Ba{(OH)_2} $ solution. pH of given hydroxide solution is:
$ A)2.699 \\
B)12.301 \\
C)7.902 \\
D)11.601 \\ $

Answer 1

Hint :pH is a proportion of how acidic/basic water is. The reach goes from $ 0to14 $ . The pH under 7 considered as acids and pH of more prominent than $ 7 $ demonstrates basic behavior, while with $ 7 $ being neutral. pH is actually a proportion of the general measure of free hydrogen and hydroxyl particles in the water.

Complete Step By Step Answer:
Firstly, we will calculate the number of moles of $ Hcl $
We know that the formula of the number of moles = $ \dfrac{{Mass}}{{Molarmass}} $ .
By this, we can calculate the molality of the solution
= $ \dfrac{{moles of solute}}{{volume in litre}} $ = $ $ $ \dfrac{{0.2}}{{20}} $ .
 $ {M_{Hcl}} = {10^{ - 2}} $
Now, we know $ 50ml $ solution is diluted in $ 250ml $ .
The new concentration will be
= $ {M_1}{V_1} = {M_2}{V_2} $ .
= $ 0.01M \times 50ml = {M_2} \times 250ml $
So the $ {M_2} $ is $ 2 \times {10^{ - 3}}{M_{}} $
When $ 40ml $ of this solution is diluted $ 20ml $ of $ Ba{(OH)_2} $ .The concentration of $ Ba{(OH)_2} $ is
= $ 40 \times 2 \times {10^{ - 3}} = 20 \times {M_3} \times 2 $
= $ $ $ {M_3} = 2 \times {10^{ - 3}}M $
The concentration of $ O{H^ - } $ will be
 $ = [O{H^ - }] = 2 \times (2 \times {10^{ - 3}}) \\
   = [O{H^ - }] = 4 \times {10^{ - 3}}M \\ $
Now, we will calculate the pH
 $ pOH = - Log[OH] = 2.397 \\
  pH = 14 - 2.397 \\
  pH = 11.6 \\ $
So, the pH of $ Ba{(OH)_2} $ IS 11.6.
Correct answer is $ D) $ .

Note :
By the use of filter paper we can find the pH of aqueous arrangement by treating it with the pH indicator. Litmus paper will be paper that has been treated with a particular indicator—a combination of 10 to 15 normal colors obtained from lichens that becomes red because of acidic conditions (pH 7). At the point when the pH is impartial (pH = 7), at that point the color is purple.