Question

How many grams of solid potassium hydroxide are needed to react with 19.0 mL of a 1.01 M nitric acid solution? Assume that the volume remains constant when the potassium hydroxide is added. KOH + HNO3KNO3 + H2O

Answer 1

we have the Balanced chemical equation as:

HNO3 + KOH ---> Na2SO4 + H2O

lets calculate the mol of HNO3

volume , V = 19.0 mL

= 1.9*10^-2 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 1.01*0.019

= 1.919*10^-2 mol

From balanced chemical reaction, we see that

when 1 mol of HNO3 reacts, 1 mol of KOH reacts

mol of KOH reacted = (1/1)* moles of HNO3

= (1/1)*1.919*10^-2

= 1.919*10^-2 mol

This is number of moles of KOH

Molar mass of KOH = 1*MM(K) + 1*MM(O) + 1*MM(H)

= 1*39.1 + 1*16.0 + 1*1.008

= 56.108 g/mol

we have below equation to be used:

mass of KOH,

m = number of mol * molar mass

= 1.919*10^-2 mol * 56.108 g/mol

= 1.08 g

Answer: 1.08 g