Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.

a. 1.3 × 10-6 M
b. 7.7 × 10-9 M
c. 1.1 × 10-10 M
d. 8.8 × 10-5 M
e. 4.9 × 10-4 M

8.8X10^-5 M

Explanation:

HClO is a weak acid that ionizes by the equation:

HClO + H2O <--> H3O+ + ClO-

Ka = [H3O+][ClO-]/[HClO] = 2.9X10^-8

Let [H3O+] = [ClO-] = x. [HClO] = 0.265 -x.

Because Ka is small, assume x << 0.265 and can be ignored. Then,

Ka = 2.9X10^-8 = x^2 / 0.265

x = [H3O+] = 8.8X10^-5 M

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