What will be the amount of (NH4)2SO4 (in g) which must be added to 500 mL of 0.2 M NH4OH to yield a solution of pH 9.35? [Given, pKa of NH4+ = 9.26, pKb NH4OH = 14 - pKa(NH4+)]

Question

What will be the amount of (NH4)2SO4 (in g) which must be added to 500 mL of 0.2 M NH4OH to yield a solution of pH 9.35? [Given, pKa of NH4+ = 9.26, pKb NH4OH = 14 - pKa(NH4+)] (A) 5.35 (B) 6.47 (C) 10.03 (D) 7.34

Tardigrade Admin · Accepted Answer

pKa of NH+4 = 9.26 Hence, pKb of NH4OH = 14 - 9.26 = 4.74 pOH = 14 - 9.35 = 4.65 Each 1 mol of (NH4)2SO4 furnishes 2 moles of NH+4 in solution. Let [(NH4)2SO4] = x mol L-1 [NH+4] = 2 x mol L-1 [NH4OH] = 0.2 mol L-1 using Henderson-Hasselbalch equation, pOH=pKb+log ([NH+4]/[NH4OH]) 4.65=4.74+log((2x/0.2)) This gives x = 0.081 mol L-1 = 0.0405 mol in 500 mL (as required) = 0.0405 × 132 = 5.35 g

Q. What will be the amount of $(N H_{4})_{2} S O_{4}$ (in g) which must be added to $500 m L$ of $0.2 M N H_{4} O H$ to yield a solution of $p H 9.35$ ? [Given, $p K_{a}$ of $N H_{4}^{+} = 9.26$ , $p K_{b} N H_{4} O H = 14 - p K_{a} (N H_{4}^{+})]$

$5.35$

$6.47$

$10.03$

$7.34$

Q. What will be the amount of (NH4​)2​SO4​ (in g) which must be added to 500mL of 0.2MNH4​OH to yield a solution of pH9.35? [Given, pKa​ of NH4+​=9.26, pKb​NH4​OH=14−pKa​(NH4+​)]

5.35

6.47

10.03

7.34

Q. What will be the amount of $(N H_{4})_{2} S O_{4}$ (in g) which must be added to $500 m L$ of $0.2 M N H_{4} O H$ to yield a solution of $p H 9.35$ ? [Given, $p K_{a}$ of $N H_{4}^{+} = 9.26$ , $p K_{b} N H_{4} O H = 14 - p K_{a} (N H_{4}^{+})]$

$5.35$

$6.47$

$10.03$

$7.34$