Question

its for 100 points give answer as early as possible....answer should be correct and with explanation

Answer 1

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

Answer 2

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• See the attached file ⤴️