The molar conductivity of acetic acid at infinite dilution is 390.7 and for 0.01 M acetic acid is 3.907 S cm2 mol-1. What is the pH of solution?

Question

The molar conductivity of acetic acid at infinite dilution is 390.7 and for 0.01 M acetic acid is 3.907 S cm2 mol-1. What is the pH of solution? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

α=(∧mc/∧mo)=(3.907/390.7)=0.01 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a530b944d30f3af46105f06676a531de-.png" /> ∴ [H⊕]=c α=0.01 × 0.01=10-4 M P H=- log (10-4)=4

Q. The molar conductivity of acetic acid at infinite dilution is $390.7$ and for $0.01 M$ acetic acid is $3.907 S c m^{2} m o l^{- 1}$ . What is the $p H$ of solution?

$α = \frac{\land _{m}^{c}}{\land _{m}^{o}} = \frac{3.907}{390.7} = 0.01$

$∴ [H^{\oplus}] = c α = 0.01 \times 0.01 = 1 0^{- 4} M$
$P H = - lo g (1 0^{- 4}) = 4$

Q. The molar conductivity of acetic acid at infinite dilution is 390.7 and for 0.01M acetic acid is 3.907Scm2mol−1. What is the pH of solution?

α=∧mo​∧mc​​=390.73.907​=0.01 ∴[H⊕]=cα=0.01×0.01=10−4M PH=−log(10−4)=4

Q. The molar conductivity of acetic acid at infinite dilution is $390.7$ and for $0.01 M$ acetic acid is $3.907 S c m^{2} m o l^{- 1}$ . What is the $p H$ of solution?

$α = \frac{\land _{m}^{c}}{\land _{m}^{o}} = \frac{3.907}{390.7} = 0.01$

$∴ [H^{\oplus}] = c α = 0.01 \times 0.01 = 1 0^{- 4} M$
$P H = - lo g (1 0^{- 4}) = 4$