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Find the pH of the following with explanation:

+6 votes
1,255 views
asked Aug 18, 2016 in Chemistry by Rahul Roy (7,955 points)
1)0.001 N NaOH

2)0.01 N Ca(OH)2

3)10^-8 N HCl

4)10^-3 M H2SO4

1 Answer

+8 votes
answered Aug 18, 2016 by vikash (21,287 points)
selected Jan 2, 2017 by sarthaks
 
Best answer

1. For a strong base, like NaOH and most of these, [OH-] is equal to the concentration of the base. So, if [NaOH] = 0.001 M, [OH-] = 0.001 M. pOH = - log[OH-], so for the first solution, pOH = 3. Since pH + pOH = 14, for that solution, pH = 11.

2. Ca(OH)2(aq) -------> Ca2+(aq) + 2OH-(aq) 
1 mole of Ca(OH)2 produce 2 mole of OH- 
0.01 M of Ca(OH)2 produce 0.02 M of OH- 

pOH = -log[OH-] 
pOH = -log(0.02) = 1.70 

pH = 14 - pOH = 14 - 1.70 = 12.3 

3. The pH of 10-8 N HCl can be calculated as follow:

[H+] = 10-8 N

As the solution is so dilute that water will also dissociate to produce Hions 

H2O → H+ + OH-

Now, The Concentration of H+ released from water is 10-7 as the pH of water is equal to 7.

Hence the total concentration of H+ = 10-7 + 10-8

[H+] = 10-7 + 0.1 × 10-7 

= 1.1 × 10-7 

pH = -log [H+]

Hence pH = 6.95 

Hence the pH of 10-8 N solution of HCl = 6.95.

4. pH = -log[H+], 
pH = -log(2 x 10^-3) = 2.69897000434

commented Dec 16, 2016 by Anuj Tiwari (10 points)
Sir how to calculate pH of 0.001N H2SO4 do we need to calculate normality into molarity.or directly we can put this value in formula because molarity and normality will give different answer.
commented Dec 16, 2016 by Abhishek Kumar (14,688 points)
First find the concentration of H+ using formula then calculate the ph

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