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Chapter 9: QAE (page 207)

Using activity coefficients correctly, find the pH of $1.0 脳 10^{- 2} MNaOH$ .

Short Answer

Expert verified

The pH of $1.0 脳 10^{- 2} MNaOH is 11.95$ .

Step by step solution

01

Concept used.

The relationship between pH and pOH is given as,

$pH + pOH = - {logK}_{w} pH + pOH = 14 A_{I^{+}} A_{OH} = K_{w} A_{H^{+}} = \frac{K_{w}}{A_{H^{-}}} pH = a - {logA}_{H^{+}} pH = - \log {[H^{+}]}_{+}$

02

Step2: Calculate the pH of 1.0×10-2 MNaOH.

Molarity of $NaOH = 1.0 脳 10^{- 2} M$

$[{OH}^{-}] = 1.0 脳 10^{- 2} M 纬_{OH} = 0.900 渭 = 0.0100 M$

Now calculate pH,

$A_{H^{+}} = \frac{K_{v}}{{IOHY}_{{OH}^{-}}} A_{H^{+}} = \frac{1.0 脳 10^{- 14}}{(1.0 脳 10^{2}) (0.900)} A_{H^{+}} = 1.11 脳 10^{- 12} pH = - \log [A_{1 +}] pH = 11.95$

The pH of $1.0 脳 10^{- 2} MNaOH = 11.95$ .

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Most popular questions from this chapter

Find the pH of 0.050 M triethylammonium bromide.

Neglecting activity coefficients, calculate the pHof $1.0 脳 10^{- 2} M KOH$

Write the chemical reaction whose equilibrium constant is

(a) $K_{a}$ for benzoic acid, $C_{6} H_{5} {CO}_{2} H$

(b) $K_{b}$ for benzoate ion, $C_{6} H_{5} {CO}_{2}^{2 -}$

(c) $K_{b}$ for aniline, $C_{6} H_{5} {NH}_{2}$

(d) $K_{a}$ for anilinium ion, $C_{6} H_{5} {NH}_{3}^{+}$

Consider a reaction mixture containing 100.0mLof 0.100mLborate buffer at $pH = {pK}_{3}$ = 9.24 . At $pH = {pK}_{a}$ , we know that $[H_{3} {BO}_{3}] = [H_{2} {BO}_{3}] = 0.0500 M$ . Suppose that a chemical reaction whose pHwe wish to control will be generating acid. To avoid changing the pH very much we do not want to generate more acid than would use up half of the $[H_{2} {BO}_{3}]$ .How many moles of acid could be generated without using up more than half of the $[H_{2} {BO}_{3}]$ ?What would bepH?

Calculate the quotient $[{HCO}_{2}^{-}] / [{HCO}_{2} H]$ at $pH 3.744$ if the ionic strength is $0.1 M$ by using the effective equilibrium constant listed for $渭 = 0.1$ in Appendix G.

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