Question

if 17.5 ml of 0.800 m hcl solution are needed to neutralize 5.00 ml of a household ammonia solution, what is the molar concentration of the ammonia? express your answer with the appropriate units.

Explanation:

Step1: Recall the neutralization formula

$M_1V_1 = M_2V_2$ where $M_1$ and $V_1$ are the molarity and volume of one solution, and $M_2$ and $V_2$ are the molarity and volume of the other solution.

Step2: Identify the given values

Let $M_1$ be the molarity of $HCl$, $V_1$ be the volume of $HCl$, $M_2$ be the molarity of ammonia and $V_2$ be the volume of ammonia. $M_1 = 0.800\ M$, $V_1=17.5\ mL$, $V_2 = 5.00\ mL$.

Step3: Rearrange the formula to solve for $M_2$

$M_2=\frac{M_1V_1}{V_2}$

Step4: Substitute the values into the formula

$M_2=\frac{0.800\ M\times17.5\ mL}{5.00\ mL}$
$M_2 = 2.8\ M$

Answer:

2.8 M