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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

Grade:

3 Answers

Arun
25763 Points
3 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
2 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
11 months 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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