A \(0.5895-\mathrm{g}\) sample of impure magnesium hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of \(0.2050 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then needs \(19.85 \mathrm{~mL}\) of \(0.1020 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percent by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

Short Answer

Expert verified
The percent by mass of magnesium hydroxide in the sample is approximately 91.33%.

Step by step solution

01

Write the balanced equations for the reactions

We need the balanced equations for the reactions: 1. Between magnesium hydroxide and HCl: \(Mg(OH)_2(aq) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)\) 2. Between excess HCl and NaOH: \(HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)\)
02

Calculate the moles of HCl used in both reactions

Moles of HCl reacted with the impure sample = Volume of HCl solution × Molarity of HCl solution \(= 100.0 \, \text{mL} \times 0.2050 \, \text{M} = 0.2050 \times 10^2 \, \mathrm{mmol} = 20.50 \, \mathrm{mmol}\) Moles of excess HCl = Volume of NaOH needed for titration × Molarity of NaOH \(= 19.85 \, \text{mL} \times 0.1020 \, \text{M} = 0.1020 \times 19.85 \, \mathrm{mmol} = 2.0257 \, \mathrm{mmol}\)
03

Calculate the moles of HCl that reacted with magnesium hydroxide

Moles of HCl that reacted with magnesium hydroxide = Total moles of HCl - Moles of excess HCl \(= 20.50 \, \mathrm{mmol} - 2.0257 \, \mathrm{mmol} = 18.4743 \, \mathrm{mmol}\)
04

Calculate the moles of magnesium hydroxide

Using the balanced equation of reaction between magnesium hydroxide and HCl, we know that 1 mole of magnesium hydroxide reacts with 2 moles of HCl. Thus: Moles of magnesium hydroxide = Moles of HCl reacting with magnesium hydroxide / 2 \(= 18.4743 \, \mathrm{mmol} / 2 = 9.23715 \, \mathrm{mmol}\)
05

Calculate the mass of magnesium hydroxide

Now, calculate the mass of magnesium hydroxide: Mass of magnesium hydroxide = Moles of magnesium hydroxide × Molar mass of magnesium hydroxide \(= 9.23715 \, \mathrm{mmol} \times 58.32 \, \mathrm{g\,mol^{-1}} = 538.383 \, \mathrm{mg}\)
06

Calculate the percentage of magnesium hydroxide in the sample

Finally, calculate the percent by mass of magnesium hydroxide in the sample: Percentage of magnesium hydroxide = (Mass of magnesium hydroxide / Mass of the sample) × 100% \(= (538.383 \, \mathrm{mg} / 589.5 \, \mathrm{mg}) \times 100\% = 91.33\%\) The percent by mass of magnesium hydroxide in the sample is approximately 91.33%.

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