Multiple Choice
What is the molar mass of a 0.350 g sample of a HA acid if it requires 50.0 mL of 0.440 M Sr(OH)2 to completely neutralize it? A is used as a place holder for the unknown nonmetal of the acid.
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14. Acids and Bases
Strong Acid Strong Base Titrations (Simplified)
Multiple Choice
How many grams of HNO3 are required to completely neutralize 110.0 mL of 0.770 M LiOH?
A
0.0847 g
B
57.0 g
C
84.7 g
D
5.34 g
Verified step by step guidance1
First, balance the chemical equation for the reaction between HNO3 and LiOH. The balanced equation is: HNO3(aq) + LiOH(aq) → LiNO3(aq) + H2O(l). This shows a 1:1 molar ratio between HNO3 and LiOH.
Calculate the moles of LiOH using its concentration and volume. Use the formula: moles = concentration (M) × volume (L). Convert 110.0 mL to liters by dividing by 1000.
Since the reaction is a 1:1 molar ratio, the moles of HNO3 required will be equal to the moles of LiOH calculated in the previous step.
Calculate the mass of HNO3 needed using the formula: mass = moles × molar mass. The molar mass of HNO3 is approximately 63.01 g/mol.
Multiply the moles of HNO3 by its molar mass to find the mass in grams. This will give you the amount of HNO3 required to completely neutralize the given amount of LiOH.
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