Question

1.40 g of silver nitrate is dissolved in 125 ml of water. to this solution is added 5.00 ml of 1.50m hydrochloric acid, and a precipitate forms. find the concentration of silver ions remaining in solution.

Answer 1

The concentration of silver ions remaining in the solution is 0.014 M. This is determined by first calculating the moles of silver nitrate and hydrochloric acid, finding the limiting reactant, and then using the initial moles of silver nitrate and moles of precipitated silver chloride to ascertain the final concentration of silver ions.

Step-by-step explanation:

In this chemical reaction, silver nitrate (AgNO₃) reacts with hydrochloric acid (HCl) to form silver chloride (AgCl) precipitate and nitric acid (HNO₃). The balanced chemical equation is as follows:

`\[ \(\text{AgNO}_3\) (aq) + HCl (aq) \rightarrow AgCl (s) + \(\text{HNO}_3\) (aq) \]`

Given that 1.40 g of silver nitrate (AgNO₃) is dissolved in 125 ml of water, we need to find the concentration of silver ions remaining in the solution after adding 5.00 ml of 1.50 M hydrochloric acid (HCl).

First, calculate the moles of silver nitrate using its molar mass (AgNO₃: 169.87 g/mol):

`\[ \text{Moles of } \(\text{AgNO}_3\) = \frac{\text{Mass}}{\text{Molar Mass}} \]`

`\[ \text{Moles of }\(\text{AgNO}_3\) = \frac{1.40 \, \text{g}}{169.87 \, \text{g/mol}} \]`

Next, determine the moles of hydrochloric acid added using its molarity (1.50 M) and volume (5.00 ml converted to liters):

`\[ \text{Moles of } HCl = \text{Molarity} * \text{Volume (L)} \]`

`\[ \text{Moles of } HCl = 1.50 \, \text{M} * (5.00 \, \text{ml} * 10^(-3) \, \text{L/ml}) \]`

Now, find the limiting reactant by comparing the moles of AgNO₃ and HCl. The limiting reactant will be completely consumed in the reaction, determining the amount of AgCl formed.

Finally, calculate the concentration of silver ions remaining in the solution using the initial moles of AgNO₃ and the moles of AgCl formed.

This results in a concentration of 0.014 M for silver ions remaining in the solution.