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

The molarity of the HCl solution is 0.2978 M

Step-by-step 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

Step-by-step 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* M_1* 24.6=1* 0.222* 33.0`

`M_1=0.30M`

Therefore, the molarity of the HCI solution is 0.30 M