Question

A 350-ml sample of 0.276 m hno3 is partially neutralized by 125 ml of 0.120 m ca(oh)2. find the concentration of nitric acid in the resulting solution.

Answer 1

To determine the concentration of nitric acid after partial neutralization, we perform stoichiometric calculations to account for reacted HNO3 and calculate the remaining concentration. The final concentration of nitric acid in the mixed solution is found to be approximately 0.140 M.

Step-by-step explanation:

To find the concentration of nitric acid in the resulting solution, we first need to carry out a stoichiometric calculation. The balanced chemical equation for the reaction between nitric acid (HNO3) and calcium hydroxide (Ca(OH)2) is:

HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O(l)

From the equation, it is clear that 2 moles of HNO3 are needed to react with 1 mole of Ca(OH)2. The initial moles of HNO3 are:

# moles HNO3 = 0.350 L × 0.276 M = 0.09660 mol

The moles of Ca(OH)2 used in the reaction are:

# moles Ca(OH)2 = 0.125 L × 0.120 M = 0.0150 mol

Since 1 mole of Ca(OH)2 reacts with 2 moles of HNO3, the moles of HNO3 reacted are:

# moles HNO3 reacted = 2 × 0.0150 mol = 0.0300 mol

Thus, the remaining moles of HNO3 after the reaction are:

# moles HNO3 remaining = 0.09660 mol - 0.0300 mol = 0.06660 mol

The total volume of the solution after mixing is:

Total volume = 0.350 L + 0.125 L = 0.475 L

Therefore, the concentration of HNO3 in the resulting solution is:

Concentration of HNO3 after reaction = # moles HNO3 remaining / Total volume = 0.06660 mol / 0.475 L = 0.140 M

Answer 2

To find the concentration of nitric acid in the resulting solution, we can use stoichiometry and the balanced chemical equation between HNO3 and Ca(OH)2. By calculating the moles of Ca(OH)2 and then deducing the moles of HNO3, we can determine the concentration of HNO3 in the solution.

Step-by-step explanation:

To find the concentration of nitric acid in the resulting solution, we can use the concept of stoichiometry and the balanced chemical equation between HNO3 and Ca(OH)2. According to the balanced equation, 2 moles of HNO3 react with 1 mole of Ca(OH)2. If 125 mL of 0.120 M Ca(OH)2 is used, we can calculate the moles of Ca(OH)2 in the solution:

moles of Ca(OH)2 = (0.125 L) x (0.120 moles/L) = 0.015 moles

Since the stoichiometry is 2:1, the moles of HNO3 in the solution will be half of the moles of Ca(OH)2:

moles of HNO3 = 0.015 moles / 2 = 0.0075 moles

To find the concentration of HNO3 in the solution, we divide the moles of HNO3 by the volume of the solution:

concentration of HNO3 = 0.0075 moles / 0.350 L = 0.02143 M