Question

Coursehero what is the concentration of the hcl solution you chose for your dilute naoh titration?

Answer 1

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 $H_{2} SO_{4}$ whose concentration is known.

Equation:

$2NaOH + H_{2} SO_{4}$  ⇒  $Na_{2} SO_{4} + 2H_{2}O$

Molar ratios:

$2mol NaOH : 1 mol H_{2} SO_{4}$

Number of moles of $H_{2} SO_{4}$ in 50.0 ml of 0.0782 M solution

$M = \frac{n}{V}\\\\n = M*V\\\\=0.0782 M * 0.050I\\\\=0.00391 mol H_{2} SO_{4}$  

Number of moles of $NaOH$

2 moles $NaOH$ / 1mol $H_{2} SO_{4}$ * 0.00391 mol $H_{2} SO_{4}$ = 0.00782 mol $NaOH$

Concentration of the solution of $NaOH$

M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M

Standardize the solution of HCl

Chemical reaction:

$NaOH + HCl$  ⇒  $NaCl + H_{2} O$

Molar ratios

1 mol $NaOH$ : 1 mol $HCl$

Number of moles of $NaOH$ in 27.5 ml

$M = \frac{n}{V} \\\\n= M*V\\\\=0.425 M * 0.0275I\\\\=0.01169 moles NaOH$

Number of moles of $HCl$

1 mol $HCl$ / 1mol $NaOH$ * 0.01169 mol $NaOH$ = 0.01169 mol $HCl$

Concentration of the solution of $HCl$

$M = \frac{n}{V}\\ \\=\frac{0.01169 mol}{0.100 l}\\ \\0.01169 M$

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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