Butyric acid is responsible for the odor in rancid butter. A solution of 0.25 m butyric acid has a ph of 2.71. What is the ka for the acid?

Ka = 1.5 -10^-5

Explanation:

Step 1: Data given

Molarity of the solution = 0.25 M

pH = 2.71

Step 2 The equation

C3H7COOH + NaOH ⇆ C3H7COONa + H2O

Step 3: Calculate pKa

pH = (pKa-log[acid])/2

2pH = pKa -log[acid]

pKa = 2pH +log[acid]

⇒with pH = 2.71

⇒with log[acid] = -0.60

pKa=2*2.71 -0.60

pKa = 5.42 - 0.60

pKa = 4.82

Step 4: Calculate Ka

pKa = -log(Ka)

Ka = 10^-4.82

Ka = 1.5 -10^-5

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2