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Chapter 22: Problem 34

In water, \(\mathrm{O}^{2-}\) is a strong base. If \(50.0 \mathrm{mg}\) of \(\mathrm{Li}_{2} \mathrm{O}\) is dissolved in \(750.0 \mathrm{mL}\) of aqueous solution, what will be the pH of the solution?

Short Answer

Expert verified
The pH of the solution is approximately 11.65.

Step by step solution

01

Calculating moles of lithium oxide

To calculate the number of moles of lithium oxide \(Li_2O\), the given mass \(50.0mg = 0.050g\) is divided by the molar mass of \(Li_2O\) which is approximately \(30g/mol\). Therefore, the moles of \(Li_2O\) is \(\frac{0.050g}{30g/mol}=0.00167mol\).
02

Calculating concentration of hydroxide ions

When Li2O dissolves in water, it forms 2OH- ions per formula unit, so the 0.00167 mol of Li2O will produce 2*0.00167=0.00333 mol of OH-. To calculate the concentration of OH- ions, the number of moles of OH- divide by the volume of the solution in liters which is 0.750L. Therefore, the concentration of OH- is \(\frac{0.00333mol}{0.75L}= 0.00444 M\).
03

Calculating the pH of the solution

To calculate the pH, we first need to find the concentration of hydronium ions \([H_3O^+]\) from the hydroxide concentration using the ion product of water, which is \( KW = [H_3O^+][OH^-] \). At 25°C, KW is \(1.00 x 10^-14 \). Therefore, \( [H_3O^+] = \frac{KW}{{OH^-}} = \frac{1.00 x 10^-14}{0.00444}= 2.25 x 10^-12 \). The pH is calculated using the formula \( pH = -log[H_3O^+] \). When the values are substituted, the pH of the solution is calculated to be about 11.65.

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Most popular questions from this chapter

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