What is the molarity of the HNO3 solution (BEFORE titration) ?

Question

A 9.54-g sample of barium hydroxide is dissolved and diluted to the mark in 300.0-ml volumetric flask. It was found that 12.80 mL of this solution was needed to reach the stoichiometric point in a titration of 26.24 mL of a solution of HNO3. What is the molarity of the HNO3 solution (BEFORE titration) ?

Answer 1

Use (m1v1) / (m2/v2) = 1/2 where m1 is the unknown molarity of the acid, m2 is the molarity of the barium salt, v1 = 26.24 and 12.8 = v2

Work out the molarity of the barium hydroxide by multiplying 9.54 by (1000/300) and dividing this by its mol wt. then do the arithmetic as described in the top line!

At least this way, you'll do some of the work yourself, and it won't look so much like plagiarism.

Answer 2

Nicole, why can't YOU answer this question?
The equation is
Ba(OH)2 + 2HNO3 -> Ba (NO3)2 + 2 H2O.
1. Find the moles in 9.54 g of Ba(OH)2.
2. Find the moles in 12.80 ml, given that all the hydroxide was added to 300 ml.
3. Compute the moles in the HNO3 solution based on the equation:
Vol x Molarity of Hydroxide = Volume x 2 Molarity of Nitric Acid . You have the 2 volumes and molarity of the hydroxide.

Answer 3

Use the equation: - (M1)(V1) = (M2)(V2) Molarity is in moles/liter (or M) and V is in liters. Convert taht 3 hundred mL to .3 L use the grams divided via molar hundreds to get Moles and you purely remedy for M2.