Answer:
[tex]\boxed{\text{(E) }1.0 \times 10^{-3} \text{ mol/L and pH = 3}}[/tex]
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
[tex]K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 1.0 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.0 \times 10^{-5}\\\\\\\text{Check for negligibility of }x\\\\\dfrac{ 0.100 }{1.0 \times 10^{-5}} = 10 000 > 400\\\\\therefore x \ll 0.100\\\\\\x^{2} = 0.100 \times 1.0\times 10^{-5} = 1.00 \times 10^{-6}\\\\x = \sqrt{1.00 \times 10^{-6}} = 1.0\times 10^{-3}\\\\\rm [H$_{3}$O$^{+}$]= x mol$\cdot$L$^{-1}$ = 1.0 $\times$ 10$^{-3}$ mol$\cdot$L$^{-1}$[/tex]
2. Calculate the pH
[tex]\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{1.0 \times 10^{-3}} = \boxed{\mathbf{3}}[/tex]
