-K-a- the acid ionisation constant of - Fe - 3- - and - H - - - is 6-5times - 10 - -3 - - What is the maximum pH value which could be used so that least 95- of total - Fe - 3- - ion in solution exists free-

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Toppr Admin · Accepted Answer

Fe3-H2O-x21CC-Fe-OH-2-H-K-Fe-OH-2-2-H-Fe3-5100-xD7-H-95-100-6-5-xD7-10-x2212-3-H-0-12MpH-x2212-log0-12-0-91

$K_{a}$ for the acid ionisation constant of ${F e}^{3 +}$ and $H^{+}$ is $6.5 \times 10^{- 3}$ . What is the maximum pH value which could be used so that at least $95$ % of total ${F e}^{3 +}$ ion in solution exists free?

${F e}^{3 +} + H_{2} O ⇌ F e {(O H)}^{2 +} + H^{+}$
$K = \frac{{[F e {(O H)}_{2}]}^{2 +} [H^{+}]}{[{F e}^{3 +}]} = \frac{\frac{5}{100} \times [H^{+}]}{95 / 100} = 6.5 \times 10^{- 3}$
$[H^{+}] = 0.12 M$
$p H = - log 0.12 = 0.91$

Ka for the acid ionisation constant of Fe3+ and H+ is 6.5×10−3. What is the maximum pH value which could be used so that at least 95% of total Fe3+ ion in solution exists free?

Fe3++H2O⇌Fe(OH)2++H+K=[Fe(OH)2]2+[H+][Fe3+]=5100×[H+]95/100=6.5×10−3[H+]=0.12MpH=−log0.12=0.91

$K_{a}$ for the acid ionisation constant of ${F e}^{3 +}$ and $H^{+}$ is $6.5 \times 10^{- 3}$ . What is the maximum pH value which could be used so that at least $95$ % of total ${F e}^{3 +}$ ion in solution exists free?

${F e}^{3 +} + H_{2} O ⇌ F e {(O H)}^{2 +} + H^{+}$
$K = \frac{{[F e {(O H)}_{2}]}^{2 +} [H^{+}]}{[{F e}^{3 +}]} = \frac{\frac{5}{100} \times [H^{+}]}{95 / 100} = 6.5 \times 10^{- 3}$
$[H^{+}] = 0.12 M$
$p H = - log 0.12 = 0.91$