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A solid sample contains both NaOH and NaCl. 0.500 g of this solid sample was dissolved in water to make a 20.0 mL solution and then this solution was titrated by 0.500 mol/L HCl solution. If 20.8 mL of HCl was used to reach the end point, what is the mass % of NaOH in the original solid sample? NaOH + HCl --> NaCl + H2O

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Explanation:

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The percent of the sample is 73.2%

Titration of HCl with NaOH:

Neutralization reactions involve the reaction of an acid and a base to produce salt (ionic compound) and water.

Acid + Base ( Salt + Water)

It is given that,

[tex]M = 0.500 g[/tex] of sample

[tex]V = 20 ml[/tex]

Solution titrated with [tex]M = 0.5 HCl[/tex]

V =[tex]18.3 ml[/tex]

Since the ratio is [tex]HCl:NaOH[/tex] (1 to 1)

Then, mol of[tex]HCl[/tex] = mol of[tex]NaOH[/tex] mol of [tex]HCl = MV =[/tex] [tex]0.5\times18.3 = 9.15[/tex] mmol of [tex]HCl[/tex]

Therefore [tex]9.15 mmol[/tex] of [tex]NaOH[/tex]

Molecular weight of [tex]NaOH = 40[/tex]

[tex]Mass = mol\times MW \\= (9.15\times10^-3)\times40 \\= 0.366 g[/tex]

So, the percent of sample = [tex]\frac{0.366}{0.5}\times 100=73.2[/tex] % is [tex]NaOH[/tex]

Learn more about the topic titration of HCl with NaOH: https://brainly.com/question/15407417

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