I. Suppose 25.0mL of a 0.265 M solution of KOH are needed to neutralize 31.0 ml.
ID: 591168 • Letter: I
Question
I. Suppose 25.0mL of a 0.265 M solution of KOH are needed to neutralize 31.0 ml.of HNO, in the reaction: HNO3 KOH KNO3 + H2O many male of K b. How many moles of HNO, must therefore have been present in the 31.0mL sample of the acid solution? answer from lb. and the 31.0 ml sample size of the HNO, solution to determine the molarity of the acid solution. 2. Suppose 1.25 grams of a diprotic acid of the general formula H,X are neutralized by 35.0ml of a 0.265 M NaOH solution in the reaction: H2X 2 NaOH Na2X +H20 How many moles of NaOH were present in the 35.0 ml of NaOH? a. b. How many moles of HzX must have reacted with the moles of NaOH calculated in 2a? What is the molar mass of the acid? (It will equal grams acid / moles acid.) c. Chemistry 111 -Concepts of Chemistry
Explanation / Answer
1. HNO3 + KOH -----------> KNO3 + H2O
1 mole 1 mole
a. no of moles of KOH = molarity * volume in L
= 0.265*0.025 = 0.006625 moles
b. from balanced equation
1 mole of KOH react with 1 mole of HNO3
0.006625 moles of KOH react with 0.006625 moles of HNO3
no of moles of HNO3 = molarity * volume in L
= 0.214*0.031 = 0.006625moles
c. HNO3 KOH
M1 = M2 = 0.265M
V1 = 31ml V2 = 25ml
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
M1 = M2V2n1/V1n2
= 0.265*25*1/31*1 = 0.214M
2.H2X + 2NaOH ------------> Na2X + H2O
a. no of moles of NaOH = molarity * volume in L
= 0.265*0.035 = 0.009275 moles
b. from balanced equation
2 moles of NaOH react with 1 mole of H2X
0.009275 moles of NaOH react with = 1*0.009275/2 = 0.0046375 moles of H2X
c. molar mass of acid = mass of acid/no of moles of acid
= 1.25/0.0046375 = 269.54g/mole