Respuesta :
The concentration of HCl is 0.117 M and the concentration of NaOH is 0.425M.
Start by standardizing the solution of NaOH by using the solution of [tex]H_{2} SO_{4}[/tex] whose concentration is known.
Equation:
[tex]2NaOH + H_{2} SO_{4}[/tex] ⇒ [tex]Na_{2} SO_{4} + 2H_{2}O[/tex]
Molar ratios:
[tex]2mol NaOH : 1 mol H_{2} SO_{4}[/tex]
Number of moles of [tex]H_{2} SO_{4}[/tex] in 50.0 ml of 0.0782 M solution
[tex]M = \frac{n}{V}\\\\n = M*V\\\\=0.0782 M * 0.050I\\\\=0.00391 mol H_{2} SO_{4}[/tex]
Number of moles of [tex]NaOH[/tex]
2 moles [tex]NaOH[/tex] / 1mol [tex]H_{2} SO_{4}[/tex] * 0.00391 mol [tex]H_{2} SO_{4}[/tex] = 0.00782 mol [tex]NaOH[/tex]
Concentration of the solution of [tex]NaOH[/tex]
M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M
Standardize the solution of HCl
Chemical reaction:
[tex]NaOH + HCl[/tex] ⇒ [tex]NaCl + H_{2} O[/tex]
Molar ratios
1 mol [tex]NaOH[/tex] : 1 mol [tex]HCl[/tex]
Number of moles of [tex]NaOH[/tex] in 27.5 ml
[tex]M = \frac{n}{V} \\\\n= M*V\\\\=0.425 M * 0.0275I\\\\=0.01169 moles NaOH[/tex]
Number of moles of [tex]HCl[/tex]
1 mol [tex]HCl[/tex] / 1mol [tex]NaOH[/tex] * 0.01169 mol [tex]NaOH[/tex] = 0.01169 mol [tex]HCl[/tex]
Concentration of the solution of [tex]HCl[/tex]
[tex]M = \frac{n}{V}\\ \\=\frac{0.01169 mol}{0.100 l}\\ \\0.01169 M[/tex]
Rounded to 3 significant figures = 0.117 M
[NaOH] = 0.425 M
[HCl] = 0.117 M
Thus the concentration of HCl and NaOH are 0.117M and 0.425M respectively.
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