Question

If it requires 48.3 milliliters of 0.55 molar nitric acid to neutralize 15.0 milliliters of barium hydroxide, solve for the molarity of the barium hydroxide solution. Show all of the work used to solve this problem. Unbalanced equation: Ba(OH)2 + HNO3 yields Ba(NO3)2 + H2O

Answer 1

First you have to moles so multiply .0483L X .55M= .026565 Multiply moles by mole ratio which is 1/2, so the moles becomes .013283 now molarity=moles/volume; divide .013283/.015L=.885533M significant figures and you final answer is 0.89M

Answer 2

Answer: The molarity of barium hydroxide solution is 0.8855 M.

Explanation:

The balanced chemical equation for the reaction is given as:

$Ba(OH)_2+2HNO_3\rightarrow Ba(NO_3)_2+2H_2O$

To calculate the molarity for a neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$

We are given:

$n_1=1\\M_1=0.55M\\V_1=48.3mL\\n_2=2\\M_2=?M\\V_2=15mL$

Putting values in above equation, we get:

$1\times 0.55\times 48.3=2\times M_2\times 15\\\\M_2=0.8855M$

Hence, the molarity of barium hydroxide solution is 0.8855 M.