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64. Calculate the pH of a solution which contains 9.9 ml of 1 M HCl and 100 ml of 0.1 M NaOH.

64. Calculate the pH of a solution which contains 9.9 ml of 1 M HCl and 100 ml of 0.1 M NaOH. 

Grade:10

1 Answers

Arun
25750 Points
6 years ago
Dear Anubhav
 
NaOH reacts with HCl : 
NaOH + HCl → NaCl + H2O 
1 mol NaOH reacts with 1 mol HCl 

Mol HCl in 9.9mL of 1.0M solution = 9.9/1000*1.0 = 0.0099 mol HCl 
Mol NaOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NaOH 
On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1*10^-4 mol NaOH dissolved in 109.9mL solution 
Molarity of NaOH solution = (1*10^-4) / 0.1099 = 9.099*10^-4M 
pOH = -log ( 9.099*10^-4) 
pOH = 3.04 
pH = 14.00 - pOH 
pH = 14.00 - 3.04 
pH = 10.96
 
Regards
Arun (askIITians forum expert)

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