PH of 0-001M solution of Zn-Cl-2- is - -Zn-2-H-2-Orightleftharpoons Zn-OH-H- Use - -K-a-1-0times -10-9-

Question

Toppr Admin · Accepted Answer

Zn2-H2O-x21CC-Zn-OH-H-xA0- -xA0- -xA0-H-c-x22C5-h-c-x221A-Kac-x221A-Ka-x22C5-c-x221A-10-x2212-9-xD7-0-001-10-x2212-6M-x2234-pH-x2212-log-H-x2212-log-10-x2212-6-6

$p H$ of $0.001 M$ solution of $Z n {C l}_{2}$ is : ${Z n}^{2 +} + H_{2} O ⇌ Z n {(O H)}^{+} + H^{+}$
Use : [ $K_{a} = 1.0 \times 10^{- 9}$ ]

${Z n}^{2 +} + H_{2} O ⇌ Z n {(O H)}^{+} + H^{+}$

$[H^{+}] = c \cdot h = c \sqrt{\frac{K_{a}}{c}} = \sqrt{K_{a} \cdot c} = \sqrt{10^{- 9} \times 0.001} = 10^{- 6} M$
$∴ p H = - l o g [H^{+}] = - l o g [10^{- 6}] = 6$

pH of 0.001M solution of ZnCl2 is : Zn2++H2O⇌Zn(OH)++H+ Use : [Ka=1.0×10−9]

Zn2++H2O⇌Zn(OH)++H+ [H+]=c⋅h=c√Kac=√Ka⋅c=√10−9×0.001=10−6M∴pH=−log[H+]=−log[10−6]=6

$p H$ of $0.001 M$ solution of $Z n {C l}_{2}$ is : ${Z n}^{2 +} + H_{2} O ⇌ Z n {(O H)}^{+} + H^{+}$
Use : [ $K_{a} = 1.0 \times 10^{- 9}$ ]

${Z n}^{2 +} + H_{2} O ⇌ Z n {(O H)}^{+} + H^{+}$

$[H^{+}] = c \cdot h = c \sqrt{\frac{K_{a}}{c}} = \sqrt{K_{a} \cdot c} = \sqrt{10^{- 9} \times 0.001} = 10^{- 6} M$
$∴ p H = - l o g [H^{+}] = - l o g [10^{- 6}] = 6$