0.01 moles of a weak acid HA K_a=2.0×10^(-6) is dissolved in 1.0 L of 0.1M HCl solution. The degree of dissociation of HA is __________ ×10^(-5)…

Question

0.01 moles of a weak acid HA K_a=2.0×10^(-6) is dissolved in 1.0 L of 0.1M HCl solution. The degree of dissociation of HA is __________ ×10^(-5) (Round off to the Nearest Integer). [Neglect volume change on adding HA and assume degree of dissociation <<1]

MARKS Content Team · Accepted Answer

Correct answer: 2. HA ⇌ H^(+) + A^(-) Initial conc. 0.1M 0M Equ. conc. (0.01-x) (0.1+x) xM ≈0.01M ≈0.1M Now, K_a=(H^(+)A^(-))/([HA])⇒2×10^(-6)=(0.1×x)/0.01 ∴x=2×10^(-7) Now, α=x/0.01=(2×10^(-7))/0.01=2×10^(-5)

$0.01$ moles of a weak acid $HA (K_{a} = 2.0 \times 10^{- 6})$ is dissolved in $1.0 L$ of $0.1 M HCl$ solution. The degree of dissociation of $HA$ is __________ $\times 10^{- 5}$ (Round off to the Nearest Integer). [Neglect volume change on adding $HA$ and assume degree of dissociation $< < 1$ ]

Solution

Relevant Questions

0.01 moles of a weak acid HA Ka=2.0×10-6 is dissolved in 1.0 L of 0.1M HCl solution. The degree of dissociation of HA is __________ ×10-5 (Round off to the Nearest Integer). [Neglect volume change on adding HA and assume degree of dissociation <<1]

Solution

Relevant Questions

$0.01$ moles of a weak acid $HA (K_{a} = 2.0 \times 10^{- 6})$ is dissolved in $1.0 L$ of $0.1 M HCl$ solution. The degree of dissociation of $HA$ is __________ $\times 10^{- 5}$ (Round off to the Nearest Integer). [Neglect volume change on adding $HA$ and assume degree of dissociation $< < 1$ ]