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100 c.c. of N/10 NaOH solution is mixed with 100 c.c. of N/5 HCl solution and the whole volume is made to 1 litre. The pH of the resulting solution will be

100 c.c. of N/10 NaOH solution is mixed with 100 c.c. of N/5 HCl solution and the whole volume is made to 1 litre. The pH of the resulting solution will be

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3 Answers

Arun
25750 Points
6 years ago
NaOH Volume = 100 ml or 0.1 L
Normality = N/10
Molarity = normality = 0.1 M
moles of NaOH = 0.1 * 0.1 = 0.01 mol
 
HCl volume = 100 ml = 0.1 L
Normality = Molarity = 0.2 M
Moles of HCl = 0.2 * 0.1 = 0.02 mol
 
0.01 mol of NaOH neutralizes 0.01 mol of HCl
as HCl is in excess, mol of HCl unreacted = 0.02 –0.01 = 0.01 mol
these free ions determine the pH of the solution
moles of HCl left after neutralization = 0.01 mol
Total volume of HCl solution = 1 L
concentration of HCl solution = moles of HCL/ volume of solution in Litres = 0.01 M
pH = –log [H+]
pH = –log (0.01) = 2
Neelam
15 Points
5 years ago
Here we use gram equivalent for acid and base and then subtract then total normality is find and ph is calculate
ankit singh
askIITians Faculty 614 Points
3 years ago
NaOH volume = 100ml or 0.1L normality =N/10 molarity=normality= 0.1M
moles of NaOH = 0.1 × 0.1 = 0.01mol [using formula mole = C (concentration in M) × V in L]
HCl volume = 100ml or 0.1L normality = molairty = 0.2M
Moles of HCl = 0.2 × 0.1 = 0.02mol
0.01 mole of NaOH neutralizes 0.01 mole of HCl
As HCl is in excess, moles of HCl left unreacted = 0.02-0.01 = 0.01mol
These free ions determine the pH of the solution
Moles of HCl left after neutralization = 0.01mol
Total volume of HCl solution = 1L
Concentration of HCl solution = moles of HCl/ volume of solution in L = 0.01M
pH = -log[H+]
pH = -(log 0.01) = -(-2) = 2
 

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