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Chapter 16: Problem 27

What mass of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), would you dissolve in \(350.0 \mathrm{mL}\) of water to produce a solution with a \(\mathrm{pH}=2.85 ?\) $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} & \\ K_{\mathrm{a}}=6.3 \times 10^{-5} \end{aligned}$$

Short Answer

Expert verified
Calculate the mass using the formula \( mass = moles * molar mass \) where moles was derived by multiplying the concentration by volume.

Step by step solution

01

Calculate pH

From the given pH=2.85, we can work out the concentration of hydronium ions \(H_{3}O^{+}\) using the formula: \( \[ [H_{3}O^{+}] = 10^{-pH} \] \). Plugging the given pH=2.85 into the formula we get: \( \[ [H_{3}O^{+}] = 10^{-2.85} \] \). Calculate this expression to find the hydronium ion concentration.
02

Determine Acid Concentration

Given the hydronium ions concentration and the acid dissociation constant \(K_{a} = 6.3 x 10^{-5}\), we can rearrange the formula for the \(K_{a}\) given by \(K_{a} = [H_{3}O^{+}][C_{6}H_{5}COO^-]/[C_{6}H_{5}COOH]\) to calculate the ratio of benzoic acid to its conjugate base. Because the acid is only weakly dissociated, we can assume that the acid concentration \([C_{6}H_{5}COOH]\) is approximately equal to the hydronium ion concentration. This gives \([C_{6}H_{5}COOH] = [H_{3}O^{+}]\). Calculate this expression to find the concentration of benzoic acid in the solution.
03

Calculate Required Mass

To find the required mass of benzoic acid, calculate the molar mass of benzoic acid first. The molar mass of benzoic acid is \(122.12 g/mol\). Knowing the concentration of benzoic acid from step 2 and the volume of the solution, we can calculate the moles of benzoic acid in the solution using the formula: \[ moles = concentration * volume \]. Subsequently, the mass can be calculated as the moles multiplied by the molar mass.

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Most popular questions from this chapter

How many milliliters of concentrated HCl(aq) \((36.0 \%\) HCl by mass, \(d=1.18 \mathrm{g} / \mathrm{mL}\) ) are required to produce \(12.5 \mathrm{L}\) of a solution with \(\mathrm{pH}=2.10 ?\)

What is the \(\mathrm{pH}\) of the solution obtained when \(125 \mathrm{mL}\) of \(0.606 \mathrm{M} \mathrm{NaOH}\) is diluted to \(15.0 \mathrm{L}\) with water?

\(50.00 \mathrm{mL}\) of \(0.0155 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) is mixed with \(75.00 \mathrm{mL}\) of 0.0106 M KOH(aq). What is the pH of the final solution?

The conjugate acid of \(\mathrm{HPO}_{4}^{2-}\) is (a) \(\mathrm{PO}_{4}^{3-}\) (b) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} ;(\mathrm{c}) \mathrm{H}_{3} \mathrm{PO}_{4} ;(\mathrm{d}) \mathrm{H}_{3} \mathrm{O}^{+} ;\) (e) none of these.

Explain why \(\left[\mathrm{PO}_{4}^{3-}\right]\) in \(1.00 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) is not simply \(\frac{1}{3}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\) but much, much less than \(\frac{1}{3}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)
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