Respuesta :
Answer: 0.30 M
Explanation:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]HCl[/tex] solution = ?
[tex]V_1[/tex] = volume of [tex]HCl[/tex] solution = 24.6 ml
[tex]M_2[/tex] = molarity of [tex]NaOH[/tex] solution = 0.222 M
[tex]V_2[/tex] = volume of [tex]NaOH[/tex] solution = 33.0 ml
[tex]n_1[/tex] = valency of [tex]HCl[/tex] = 21
[tex]n_2[/tex] = valency of [tex]NaOH[/tex] = 1
[tex]1\times M_1\times 24.6=1\times 0.222\times 33.0[/tex]
[tex]M_1=0.30M[/tex]
Therefore, the molarity of the HCI solution is 0.30 M
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
