Respuesta :
Answer:
Molarity of acetic acid is 0.812 M
Explanation:
Balanced reaction: [tex]2HC_{2}H_{3}O_{2}+Ca(OH)_{2}\rightleftharpoons Ca(C_{2}H_{3}O_{2})_{2}+2H_{2}O[/tex]
According to balanced equation, 1 mol of [tex]Ca(OH)_{2}[/tex] neutralizes 2 moles of [tex]HC_{2}H_{3}O_{2}[/tex]
Number of moles of [tex]Ca(OH)_{2}[/tex] in 33.93 mL of 0.0599 M [tex]Ca(OH)_{2}[/tex] = [tex]\frac{0.0599\times 33.93}{1000}moles=0.00203moles[/tex]
Let's assume molarity of [tex]HC_{2}H_{3}O_{2}[/tex] is C (M)
Then, number of moles acetic acid were neutralized = [tex]\frac{5.00\times C}{1000}moles=0.005C[/tex]
So, [tex]0.005C=2\times 0.00203[/tex]
or, [tex]C=0.812[/tex]
So, molarity of acetic acid is 0.812 M
The Molarity of acetic acid is 0.812 M.
Calculation of the molarity:
But before that first we have to determine the number of moles
= 0.0599 * 33.93 / 1000
= 0.00203 miles
Now the molarity should be
= 5.00 * C/1000 = 0.005C
So,
0.005C = 2 * 0.00203
So, C = 0.812 M
hence, we can conclude that The Molarity of acetic acid is 0.812 M.
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