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Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
Let empirical formula for hydrocarbon is CxHy
it will undergo combustion as
CxHy + (x + y/4) O2 ---> xCO2 + (y/2 )H2O
Given that mass of CO2 produced = 9.69 g
So moles of CO2 produced = 9.69 / 44 = 0.22 moles
So moles of carbon present = 0.22 moles
mass of H2O produced = 4.96 g
Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles
So moles of H present = 2 X 0.28 = 0.56 moles
Let us divided the moles of each with lowest value of moles
Moles of Carbon = 0.22 / 0.22 = 1 moles
moles of H = 0.56 / 0.22 = 2.55
Multiplying with two to get whole number
the ratio of carbon and hydrogen will be : C:H = 2:5
empirical formula : C2H5
