Ka for the ionisation of Fe3- to Fe OH2- and H- is 6-5- 10-3 what is the maximum pH value which could be used so that at least 95

Fe^3++H_2O nbsp;⇌ nbsp;Fe(OH)^2+ + H^+; K_h = 6.5 × 10^ 3 (1 h) [H^⊕] = Ch and K_h = C· h^2/1 h and h=5/100 Thus, 6.5× 10^ 3 = C× ...

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Byju's Answer

Standard XII

Chemistry

Introduction and Degree of Hydrolysis

Ka for the io...

Question

$K_{a}$ for the ionisation of $F e^{3 +}$ to $F e (O H)^{2 +}$ and $H^{\oplus}$ is $6.5 \times 10^{- 3}$ what is the maximum pH value which could be used so that at least 95% of the total $F e^{3 +}$ in a dilute solution exists as $F e^{3 +}$ (write your answer to nearest integer)?

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Solution

$F e^{3 +} + H_{2} O ⇌ F e (O H)^{2 +} + H^{+}; K_{h} = 6.5 \times 10^{- 3}$
(1 - h)
$[H^{\oplus}] = C h$
and $K_{h} = \frac{C \cdot h^{2}}{1 - h}$ and $h = \frac{5}{100}$
Thus, $6.5 \times 10^{- 3} = \frac{C \times 5 \times 5 \times 100}{100 \times 100 \times 95}$
$∴ C = 2.47 M ∴ [H^{\oplus}] = 2.47 \times \frac{5}{100}$
$∴ p H = 0.9083$

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Ka for the ionisation of Fe3+ to Fe(OH)2+ and H⊕ is 6.5×10−3 what is the maximum pH value which could be used so that at least 95% of the total Fe3+ in a dilute solution exists as Fe3+ (write your answer to nearest integer)?

Fe3++H2O⇌Fe(OH)2++H+;Kh=6.5×10−3(1 - h)[H⊕]=Chand Kh=C⋅h21−h and h=5100Thus, 6.5×10−3=C×5×5×100100×100×95∴C=2.47M∴[H⊕]=2.47×5100∴pH=0.9083

$K_{a}$ for the ionisation of $F e^{3 +}$ to $F e (O H)^{2 +}$ and $H^{\oplus}$ is $6.5 \times 10^{- 3}$ what is the maximum pH value which could be used so that at least 95% of the total $F e^{3 +}$ in a dilute solution exists as $F e^{3 +}$ (write your answer to nearest integer)?