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Find the \(\mathrm{pH}\) of solutions with the following \(\left[\mathrm{H}^{+}\right]\). Classify each as acidic or basic. (a) \(2.7 \times 10^{-3} \mathrm{M}\) (b) \(1.5 \mathrm{M}\) (c) \(1.45 \times 10^{-13} \mathrm{M}\) (d) \(6.4 \times 10^{-9} \mathrm{M}\)

Short Answer

Expert verified
Solution: (a) Solution a has a \(\mathrm{pH}\) value of approximately 2.57 and is classified as acidic. (b) Solution b has a \(\mathrm{pH}\) value of approximately -0.18 and is classified as acidic. (c) Solution c has a \(\mathrm{pH}\) value of approximately 12.84 and is classified as basic. (d) Solution d has a \(\mathrm{pH}\) value of approximately 8.19 and is classified as basic.

Step by step solution

01

Find the pH of each solution

To find the pH of each solution, we'll use the formula for \(\mathrm{pH}\) as mentioned before: \(\mathrm{pH} = -\log_{10}\left[\mathrm{H}^{+}\right]\) (a) For the first solution, the concentration of hydrogen ions is \(2.7 \times 10^{-3} M\). Calculate its pH: \(\mathrm{pH}_a = -\log_{10}(2.7 \times 10^{-3})\) (b) For the second solution, the concentration of hydrogen ions is \(1.5M\). Calculate its pH: \(\mathrm{pH}_b = -\log_{10}(1.5)\) (c) For the third solution, the concentration of hydrogen ions is \(1.45 \times 10^{-13} M\). Calculate its pH: \(\mathrm{pH}_c = -\log_{10}(1.45 \times 10^{-13})\) (d) For the fourth solution, the concentration of hydrogen ions is \(6.4 \times 10^{-9} M\). Calculate its pH: \(\mathrm{pH}_d = -\log_{10}(6.4 \times 10^{-9})\)
02

Evaluate the pH values

Calculate the \(\mathrm{pH}\) values for each of the solutions: (a) \(\mathrm{pH}_a = -\log_{10}(2.7 \times 10^{-3}) \approx 2.57\) (b) \(\mathrm{pH}_b = -\log_{10}(1.5) \approx -0.18\) (c) \(\mathrm{pH}_c = -\log_{10}(1.45 \times 10^{-13}) \approx 12.84\) (d) \(\mathrm{pH}_d = -\log_{10}(6.4 \times 10^{-9}) \approx 8.19\)
03

Classify each solution as acidic, basic, or neutral

Now that we have our \(\mathrm{pH}\) values, we can classify the solutions as acidic, basic, or neutral based on their \(\mathrm{pH}\): (a) \(2.57 < 7\): The first solution is acidic. (b) \(-0.18 < 7\): The second solution is acidic. (c) \(12.84 > 7\): The third solution is basic. (d) \(8.19 > 7\): The fourth solution is basic.

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Most popular questions from this chapter

Solution A has a pH of 12.32. Solution B has [H \(^{+}\) ] three times as large as that of solution A. Solution C has a pH half that of solution \(\mathrm{A}\). (a) What is \(\left[\mathrm{H}^{+}\right]\) for all three solutions? (b) What is the pH of solutions \(\mathrm{B}\) and \(\mathrm{C}\) ? (c) Classify each solution as acidic, basic, or neutral.

Find \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of the following solutions. (a) Thirty-eight \(\mathrm{mL}\) of a \(0.106 \mathrm{M}\) solution of \(\mathrm{Sr}(\mathrm{OH})_{2}\), diluted with enough water to make \(275 \mathrm{~mL}\) of solution. (b) A solution prepared by dissolving \(5.00 \mathrm{~g}\) of \(\mathrm{KOH}\) in enough water to make \(447 \mathrm{~mL}\) of solution.

Calculate \(\mathrm{p} K_{\mathrm{a}}\) for the weak acids that have the following \(K_{\mathrm{a}}\) values: (a) \(1.8 \times 10^{-4}\) (b) \(6.8 \times 10^{-8}\) (c) \(4.0 \times 10^{-11}\)

Caproic acid, \(\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2},\) is found in coconut oil and is used in making artificial flavors. A \(2.00-\mathrm{L}\) solution of caproic acid has a pH of 2.77 and contains \(52.3 \mathrm{~g}\) of caproic acid. What is \(K_{\mathrm{a}}\) for caproic acid?

Consider the process \(\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad \Delta H^{\circ}=55.8 \mathrm{~kJ}\) (a) Will the \(\mathrm{pH}\) of pure water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) be \(7.0 ?\) (b) If not, calculate the \(\mathrm{pH}\) of pure water at \(37^{\circ} \mathrm{C}\).

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