Answer:
Ethyl ethanoate is formed
Answer:
7.22 L
Explanation:
From the question,
Applying Charles law,
V₁/T₁ = V₂/T₂...................... Equation 1
Where V₁ = Initial volume of gas, V₂ = Final volume of gas, T₁ = Initial Temperature of gas in Kelvin, T₂ = Final Temperature of gas in Kelvin
Make V₂ the subject of the equation
V₂ = (V₁×T₂)/T₁................. Equation 2
Given: V₁ = 5.00 L, T₁ = 20 °C = (20+273) K = 293 K, T₂ = 150 °C = (150+273) K = 423 K
Substitute these value into equation 2
V₂ = (5.00×423)/293
V₂ = 7.22 L
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First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
Muddy water is indeed a mixture