Question

Calculate the amount of (nh4)2so4 in grams which must ve added to 500 ml of 0.2 m nh3 to give a solution of ph =9?3 .Given pka for nh3=4.7

Answer 1

Concept:

According to henderson hasselbalch equation, pOH of a solution containing salt of base and base can be calculated by following equation

$pOH = pK_{b} + log\frac{moles of salt}{moles of base}$

Explanation :

Given that,

Moles of ammonia (base) = 0.5 L × 0.2 M =  0.1 moles

pOH = 14-pH = 14-9.3 = 4.7

On substituting all these values in above formula

We get,

$4.7 =  4.7 + log\frac{moles of salt}{0.1}$

log(moles of salt) - log(0.1) = 4.7-4.7 = 0

log(moles of salt)  + 1 = 0

log(moles of salt) = -1

moles of salt = $10^{-1}$ = 0.1

Thus, amount of $(NH_{4} _){2}SO_{4}$ = 0.1×132.13 = 13.213 gram

Conclusion:

Amount of ammonium sulfate = 13.213 g