Question

What volume (in mL) of 0.09216 M HNO3 is required to react completely with 0.2419 g of potassium hydrogen phosphate? 2 HNO3(aq) + K2HPO4(aq) → H2PO4(aq) + 2 KNO3(aq)

Answer 1

Molar mass of K2HPO4,

MM = 2*MM(K) + 1*MM(H) + 1*MM(P) + 4*MM(O)

= 2*39.1 + 1*1.008 + 1*30.97 + 4*16.0

= 174.178 g/mol

mass(K2HPO4)= 0.2419 g

use:

number of mol of K2HPO4,

n = mass of K2HPO4/molar mass of K2HPO4

=(0.2419 g)/(174.178 g/mol)

= 1.389*10^-3 mol

According to balanced equation

mol of HNO3 reacted = (2/1)* moles of K2HPO4

= (2/1)*1.389*10^-3

= 2.778*10^-3 mol

This is number of moles of HNO3

use:

M = number of mol / volume in L

0.09216 = 2.778*10^-3/ volume in L

volume = 0.03014 L

volume = 30.14 mL

Answer: 30.1 mL