Respuesta :
Answer:
0.00244 moles of HCl present in the solution.
Explanation:
The balanced reaction between KOH and HCl is as follows.
[tex]HCl(aq)+KOH(aq)\rightarrow KCl(aq)+H_{2}O[/tex]
The following formula we can use to identify the number of moles of solution.
[tex]\frac{M_{1}V_{1}}{n_{1}}=\frac{M_{2}V_{2}}{n_{2}}[/tex]
Rewrite the formula is as follows:
[tex]M_{1}=\frac{M_{2}V_{2}}{n_{2}}\times \frac{n_{1}}{V_{1}}....................(1)[/tex]
From the given,
HCl:
[tex]M_{1} =?\\V_{1}\,=15\,ml\\n_{1}=1\,mol[/tex]
KOH:
[tex]M_{2} =0.252M\\V_{2}\,=9.71\,ml\\n_{2}=1\,mol[/tex]
Substitute the all values in equation (1)
[tex]M_{1}=\frac{(0.252\,M)(9.71\,ml)}{1\,mol}\times \frac{1\,mol}{15\,ml}=0.1631\,mol/L[/tex]
Therefore, moles of HCl = 0.1631 mol/L
Volume = 15 ml = 0.015 L
[tex]M=\frac{n}{V}\\n=M\times V\\=0.1631\times 0.015L =0.00244\,mol\,of\,HCl[/tex]
