A cation exchange resin RH 2 can replace Ca 2- in hard water as RH 2- Ca 2- RCa -2 H - -1 L of hard water hardness x ppm after passing through RH 2 has pH -2- Then value of x -40 is

PH=2 ⇒ [H^+] = 10^ 2 [Ca^2+]=10^ 22=5× 10^ 3 moles 5 × 10^ 3× 40 = 200 mg x=200 ppm Then, the value of x40=20040=5 ...

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Valency and Oxidation State

A cation exch...

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A cation exchange resin $R H_{2}$ can replace $C a^{2 +}$ in hard water as $R H_{2} + C a^{2 +} \to R C a + 2 H^{+}$ . $1 L$ of hard water (hardness $x p p m$ ) after passing through $R H_{2}$ has $p H = 2$ . Then value of $\frac{x}{40}$ is:

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R H_{2}

C a^{2 +}

R H_{2} + C a^{2 +} \to R C a + 2 H^{+}

1 L

R H_{2}

p H = 2

C a^{2 +}

$R H_{2}$ (ion exchange resin) can replace $C a^{2 +}$ ions in hard water as: $R H_{2} + C a^{2 +} \to R C a + 2 H^{+}$ .

If $1 L$ of hard water after passing through $R H_{2}$ has $p H = 3$ , then hardness in parts per million of $C a^{2 +}$ is:

200

H^{+}

C a^{2 +}

50

0.1 M N a O H

C a^{2 +}

C a^{2 +}

C a^{2 +}

H^{+}

C a = 40]

77.5

C a^{2 +}

C a^{2 +}

H^{+}

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