Answer:
Explanation:
We shall use the formula S₁V₁ = S₂V₂
S₁ = .1180 M , V₁ = 25.98 mL
S₂ = ? , V₂ = 52.50 mL
.1180 M x 25.98 = 52.50 x S₂
S₂ = .0584 M
Molarity of the acid solution = .0584 M .
The concentration of the acid solution is 0.058 M.
The equation of the reaction is;
CH3COOH(aq) + KOH(aq) -----> CH3COOK(aq) + H2O(l)
The following are known;
Concentration of base CB = 0.1180 M
Volume of base VB = 25.98 mL
Concentration of acid CA = ?
Volume of acid VA = 52.50 mL
Number of moles of acid NA = 1
Number of moles of base NB = 1
Using;
CAVA/CBVB =NA/NB
CAVANB = CBVBNA
CA= CBVBNA/VANB
CA = 0.1180 M × 25.98 mL × 1/52.50 mL × 1
CA = 0.058 M
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