Question

How do you do this question?

Answer 1

`\boxed{\text{(E) }1.0 * 10^(-3) \text{ mol/L and pH = 3}}`

Step-by-step explanation:

1. Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

HA + H₂O ⇌ H₃O⁺ + A⁻

I/mol·L⁻¹: 0.100 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.100 - x x x

`K_{\text{a}} = \frac{\text{[H}_(3)\text{O}^(+)]\text{A}^(-)]} {\text{[HA]}} = 1.0 * 10^(-5)\\\\(x^(2))/(0.100 - x) = 1.0 * 10^(-5)\\\\\\\text{Check for negligibility of }x\\\\( 0.100 )/(1.0 * 10^(-5)) = 10 000 > 400\\\\\therefore x \ll 0.100\\\\\\x^(2) = 0.100 * 1.0* 10^(-5) = 1.00 * 10^(-6)\\\\x = \sqrt{1.00 * 10^(-6)} = 1.0* 10^(-3)\\\\\rm [H$_(3)$O$^(+)$]= x mol$\cdot$L$^(-1)$ = 1.0 $*$ 10$^(-3)$ mol$\cdot$L$^(-1)$`

2. Calculate the pH

`\text{pH} = -\log{\rm[H_(3)O^(+)]} = -\log{1.0 * 10^(-3)} = \boxed{\mathbf{3}}`