Answer:
[tex]M_{acid}=0.08892M[/tex]
Explanation:
Hello!
In this case, since the reaction between arsenic acid and potassium hydroxide is:
[tex]H_3AsO_4+3KOH\rightarrow K_3AsO_4+3H_2O[/tex]
Thus, since the mole ratio between the acid and the base is 1:3, at the equivalence point we can write:
[tex]3n_{acid}=n_{base}[/tex]
That in terms of molarities and volumes is:
[tex]3M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, the concentration of the arsenic acid is:
[tex]M_{acid}=\frac{M_{base}V_{base}}{3V_{acid}}=\frac{0.1894M*35.21mL}{3*25.00mL} \\\\M_{acid}=0.08892M[/tex]
Best regards!
The concentration (M) of arsenic acid in a solution of 25.00 mL is 0.088M.
Molarity of any solution will be calculated as:
M = n/V, where
V = volume
n is the moles and it will be calculated as:
n = W/M, where
W = given mass
M = molar mass
Given chemical reaction is:
H₃AsO₄ + 3KOH → K₃AsO₄ + 3H₂O
Moles of 35.21 mL of 0.1894 M KOH will be calculated as:
n = (0.1894)(0.035) = 0.0066 moles
From the stoichiometry of the reaction it is clear that:
1 mole of H₃AsO₄ = react with 3 moles of KOH
0.0066 moles of KOH = react with 1/3×0.0066= 0.0022 moles of H₃AsO₄
Now we calculate the molarity of H₃AsO₄ in 25mL of solution as:
M = 0.0022/0.025 = 0.088 M
Hence, required molarity is 0.088 M.
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