The concentration of the HCl solution is 0.047 M.
Explanation:
Data given about acid and base:
volume of acid Vacid = 46.9 ml
molarity of acid =?
volume of the base (NaOH) = 16.4 ml
molarity of the base = 0.135 M
To know the concentration of the acid in this reaction, the formula used in titration is used. It is
Macid X Vacid = Mbase X Vbase
the formula is rewritten as:
Macid = 
putting the values in the equation:
Macid = 
= 0.047 M
the concentration of the acid i.e HCl in the solution is of 0.047 M.
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.
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