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Chapter 16: Problem 40

Complete the following table by calculating the missing entries. In each case indicate whether the solution is acidic or basic. $$ \begin{array}{lllll} \hline & & & & \text { Acidic or } \\ \mathrm{pH} & \mathrm{pOH} & {\left[\mathrm{H}^{+}\right]} & {\left[\mathrm{OH}^{-}\right]} & \text {basic? } \\ \hline 11.25 & & & & \\ & 6.02 & & \\ & & 4.4 \times 10^{-4} \mathrm{M} & & \\ & & & 8.5 \times 10^{-3} \mathrm{M} & \\ \hline \end{array} $$

Short Answer

Expert verified
} \\\ \hline 11.25 & 2.75 & 5.62 \times 10^{-12} M & 1.78 \times 10^{-3} M & \text { basic } \\\ 7.98 & 6.02 & 1.05 \times 10^{-8} M & 9.33 \times 10^{-7} M & \text { basic } \\\ 3.36 & 10.64 & 4.4 \times 10^{-4} M & 2.27 \times 10^{-11} M & \text { acidic } \\\ 11.93 & 2.07 & 1.18 \times 10^{-12} M & 8.5 \times 10^{-3} M & \text { basic } \\\ \hline \end{array} $$

Step by step solution

01

First Row: Finding pOH, [H+] and [OH-]

Given: pH = 11.25 Using relationship 3, pOH = 14 - pH = 14 - 11.25 = 2.75 Now, we will find [H+] and [OH-] values using relationships 1 and 2: [H+] = 10^(-pH) = 10^(-11.25) = 5.62 脳 10^(-12) M [OH-] = 10^(-pOH) = 10^(-2.75) = 1.78 脳 10^(-3) M Since pH > 7, the solution is basic.
02

Second Row: Finding pH, [H+] and [OH-]

Given: pOH = 6.02 Using relationship 3, pH = 14 - pOH = 14 - 6.02 = 7.98 Now, we will find [H+] and [OH-] values using relationships 1 and 2: [H+] = 10^(-pH) = 10^(-7.98) = 1.05 脳 10^(-8) M [OH-] = 10^(-pOH) = 10^(-6.02) = 9.33 脳 10^(-7) M Since pH > 7, the solution is basic.
03

Third Row: Finding pH, pOH and [OH-]

Given: [H+] = 4.4 脳 10^(-4) M Using relationship 1, pH = -log[H+] = -log(4.4 脳 10^(-4)) = 3.36 Using relationship 3, pOH = 14 - pH = 14 - 3.36 = 10.64 Now, we will find the [OH-] value using relationship 4: [OH-] = (1 脳 10^(-14) M虏) / [H+] = (1 脳 10^(-14) M虏) / (4.4 脳 10^(-4) M) = 2.27 脳 10^(-11) M Since pH < 7, the solution is acidic.
04

Fourth Row: Finding pH, pOH and [H+]

Given: [OH-] = 8.5 脳 10^(-3) M Using relationship 2, pOH = -log[OH-] = -log(8.5 脳 10^(-3)) = 2.07 Using relationship 3, pH = 14 - pOH = 14 - 2.07 = 11.93 Now, we will find the [H+] value using relationship 4: [H+] = (1 脳 10^(-14) M虏) / [OH-] = (1 脳 10^(-14) M虏) / (8.5 脳 10^(-3) M) = 1.18 脳 10^(-12) M Since pH > 7, the solution is basic. The completed table is: $$ \begin{array}{lllll} \hline & & & & \text { Acidic or } \\\ \mathrm{pH} & \mathrm{pOH} & {\left[\mathrm{H}^{+}\right]} & {\left[\mathrm{OH}^{-}\right]} & \text {basic? } \\\ \hline 11.25 & 2.75 & 5.62 \times 10^{-12} M & 1.78 \times 10^{-3} M & \text { basic } \\\ 7.98 & 6.02 & 1.05 \times 10^{-8} M & 9.33 \times 10^{-7} M & \text { basic } \\\ 3.36 & 10.64 & 4.4 \times 10^{-4} M & 2.27 \times 10^{-11} M & \text { acidic } \\\ 11.93 & 2.07 & 1.18 \times 10^{-12} M & 8.5 \times 10^{-3} M & \text { basic } \\\ \hline \end{array} $$

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Most popular questions from this chapter

The odor of fish is due primarily to amines, especially methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\). Fish is often served with a wedge of lemon, which contains citric acid. The amine and the acid react forming a product with no odor, thereby making the less-than-fresh fish more appetizing, Using data from Appendix D, calculate the equilibrium constant for the reaction of citric acid with methylamine, if only the first proton of the citric acid \(\left(K_{a 1}\right)\) is important in the neutralization reaction.

If a substance is an Arrhenius base, is it necessarily a Bronsted-Lowry base? Is it necessarily a Lewis base? Explain.

Explain the following observations: (a) \(\mathrm{HCl}\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S} ;\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is a stronger acid than \(\mathrm{H}_{3} \mathrm{As} \mathrm{O}_{4}\); (c) \(\mathrm{HBrO}_{3}\) is a stronger acid than \(\mathrm{HBr} \mathrm{O}_{2}\); (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is a stronger acid than \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} ;(\mathrm{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is a stronger acid than phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) ).

(a) What is a strong base? (b) A solution is labeled \(0.035 \mathrm{M} \mathrm{Sr}(\mathrm{O} \mathrm{H})_{2}\). What is \(\left[\mathrm{OH}^{-}\right]\) for the solution? (c) Is the following statement true or false? Because \(\mathrm{Mg}(\mathrm{OH})_{2}\) is not very soluble, it cannot be a strong base. Explain.

(a) Given that \(K_{a}\) for acetic acid is \(1.8 \times 10^{-5}\) and that for hypochlorous acid is \(3.0 \times 10^{-8}\), which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypochlorite ion? (c) Calculate \(K_{b}\) values for \(\mathrm{CH}_{2} \mathrm{COO}^{-}\) and \(\mathrm{Cl} \mathrm{O}^{-}\)
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