Question

write a balanced equation to determine the molarity of the HCI solution when a 24.6 ml sample of HCI reacts with a 33.0 mL of 0.222 M NaOH solution

Answer 1

Answer:

The molarity of the HCl solution is 0.2978 M

Explanation:

Step 1: Balance the reaction:

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

⇒We can see that for 1 mole reacting, we will also have 1 mole NaOH reacting ( since the ratio is 1:1).

⇒ As well as there will be produced 1 mole H2O and 1 mole of NaCL.

Step 2: Calculating moles

We have to find number of moles HCl, since we know that for 1 mole NaOH there will be 1 mole HCl ⇒ we can calculate number of moles of NaOH

⇒ we use the formule Concentration = mole / volume

moles NaOH = 0.222 M * 33.0 * 10^-3 L =0.007326 mole NaOH

⇒ We have also 0.007326 mole HCl

Step 3: Calculating molarity of HCl

⇒ Molarity is number of mole per volume

⇒ M(HCl) = 0.007326 mole HCl / (24.6 *10^-3 L)

M(HCl) =0.2978 M

The molarity of the HCl solution is 0.2978 M

Answer 2

Answer: 0.30 M

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HCl$ solution = ?

$V_1$ = volume of $HCl$ solution = 24.6 ml

$M_2$ = molarity of $NaOH$ solution = 0.222 M

$V_2$ = volume of $NaOH$ solution = 33.0 ml

$n_1$ = valency of $HCl$ = 21

$n_2$ = valency of $NaOH$ = 1

$1\times M_1\times 24.6=1\times 0.222\times 33.0$

$M_1=0.30M$

Therefore, the molarity of the HCI solution is 0.30 M