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1)0.001 N NaOH

2)0.01 N Ca(OH)2

3)10^-8 N HCl

4)10^-3 M H2SO4

1 Answer

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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

by (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.
by (9.9k points)
First find the concentration of H+ using formula then calculate the ph

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