Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
Answer:
A: NaOH + HCl = NaCl +H2O
B: Volume of NaOH = 22.2ml
C:
D: Volume of HCl = 9.45 ml
E:
Explanation:
Part A: Balanced chemical equaion
NaOH + HCl = NaCl +H2O
part B:
Volume of NaOH used in titration is 22.2 ml because volume is taken upto the end point of the titration.
Part C:
Calculation of moles of NaOH used:
Part D:
Volume of HCl used in titration is 9.45 ml because volume is taken upto the end point of the titration.
Part E:
But in case of acid it is known as basicity and in case of base it is known as acidity.
in case of NaOH and HCl n-factor is 1;
hence
Normality=molarity;