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Chapter 16: Problem 43

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: \((\mathbf{a}) 8.5 \times 10^{-3} \mathrm{M} \mathrm{HBr},(\mathbf{b}) 1.52 \mathrm{g}\) of \(\mathrm{HNO}_{3}\) in 575 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 5.00 \mathrm{mL}\) of 0.250 \(\mathrm{M} \mathrm{ClO}_{4}\) diluted to 50.0 \(\mathrm{mL}\) (d) a solution formed by mixing 10.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HBr}\) with 20.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HCl} .\)

Short Answer

Expert verified
The pH values for the strong acid solutions are as follows: (a) \(\mathrm{pH} = 2.07\) (b) \(\mathrm{pH} = 1.38\) (c) \(\mathrm{pH} = 1.60\) (d) \(\mathrm{pH} = 0.78\)

Step by step solution

01

Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

HBr is a strong acid. Therefore, it will dissociate completely into \(\mathrm{H^{+} (aq)}\) and \(\mathrm{Br^{-} (aq)}\) ions in solution. The concentration of \(\mathrm{H^{+}}\) ions is the same as the concentration of HBr, which is \(8.5 \times 10^{-3} \mathrm{M}\).
02

Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(8.5 \times 10^{-3})\). Calculate the pH: \[\mathrm{pH} = 2.07\] #b) Calculate the pH of 1.52 g of HNO₃ in 575 mL of solution:
03

Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

First, find the moles of HNO₃ by dividing the mass by the molar mass (63.01 g/mol): \(\frac{1.52\,\mathrm{g}}{63.01\, \mathrm{g/mol}} = 0.0241\,\mathrm{mol}\) Next, convert the volume of solution to liters: \(\frac{575\,\mathrm{mL}}{1000} = 0.575\,\mathrm{L}\) Now, calculate the molar concentration of HNO₃: \(\frac{0.0241\,\mathrm{mol}}{0.575\,\mathrm{L}} = 0.0419\,\mathrm{M}\) Since HNO₃ is a strong acid, the concentration of \(\mathrm{H^{+}}\) ions in the solution is the same as the concentration of HNO₃, which is \(0.0419\,\mathrm{M}\).
04

Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(0.0419)\). Calculate the pH: \[\mathrm{pH} = 1.38\] #c) Calculate the pH of 5.00 mL of 0.250 M ClO₄⁻ diluted to 50.0 mL:
05

Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

First, find the moles of HClO₄ in 5.00 mL of 0.250 M solution: \((0.250\,\mathrm{M})(0.00500\,\mathrm{L}) = 0.00125\,\mathrm{mol}\) Next, calculate the new concentration after dilution: \(\frac{0.00125\,\mathrm{mol}}{0.0500 \,\mathrm{L}} = 0.0250\,\mathrm{M}\) Since HClO₄ is a strong acid, the concentration of \(\mathrm{H^{+}}\) ions is the same as the concentration of HClO₄, which is \(0.0250\,\mathrm{M}\).
06

Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(0.0250)\). Calculate the pH: \[\mathrm{pH} = 1.60\] #d) Calculate the pH of a solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl:
07

Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

First, calculate the moles of \(\mathrm{H^+}\) ions from HBr and HCl: HBr: \((0.100\,\mathrm{M})(0.0100\,\mathrm{L}) = 0.00100\,\mathrm{mol}\) HCl: \((0.200\,\mathrm{M})(0.0200\,\mathrm{L}) = 0.00400\,\mathrm{mol}\) Next, add the moles of \(\mathrm{H^+}\) ions: \(0.00100\,\mathrm{mol} + 0.00400\,\mathrm{mol} = 0.00500\,\mathrm{mol}\) Now, calculate the total volume of the solution: \(10.0\,\mathrm{mL} + 20.0\,\mathrm{mL} = 30.0\,\mathrm{mL} = 0.0300\,\mathrm{L}\) Calculate the new concentration of \(\mathrm{H^+}\) ions: \(\frac{0.00500\,\mathrm{mol}}{0.0300\,\mathrm{L}} = 0.167\,\mathrm{M}\)
08

Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(0.167)\). Calculate the pH: \[\mathrm{pH} = 0.78\]

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Most popular questions from this chapter

The \(\mathrm{p} K_{\mathrm{b}}\) of water is ______ . (a) 1 (b) 7 (c) 14 (d) not defined (e) none of the above

Determine whether each of the following is true or false: (a) All strong bases are salts of the hydroxide ion. (b) The addition of a strong base to water produces a solution of \(\mathrm{pH}>\)7.0 .(c) Because \(\mathrm{Mg}(\mathrm{OH})_{2}\) is not very soluble, it cannot be a strong base.

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (B) \(\mathrm{HCO}_{3}^{-}\)

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-},(\mathbf{b}) \mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-}\) \((\mathbf{c}) \mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}.\)

Ammonia, \(\mathrm{NH}_{3},\) acts as an Arrhenius base, a Bronsted-Lowry base, and a Lewis base, in aqueous solution. Write out the reaction \(\mathrm{NH}_{3}\) undergoes with water and explain what properties of ammonia correspond to each of the three definitions of "base."
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