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Chapter 10: Q10AE (page 229)

Find the $pH$ and the concentrations of $H_{2} {SO}_{3,} {HSO}_{3}^{-}$ and role="math" localid="1654855262945" $S {O_{3}}^{2 -}$ in each solution:
(c) $0.050 {MNa}_{2} {SO}_{3}$

Short Answer

Expert verified

The concentration of solution is $7.2 脳 10^{- 13} M$

Step by step solution

01

Define the formula to find ph of solution.

The $pH$ of solution can be calculated as

$[H^{+}] = \sqrt{\frac{K_{1} K_{2} (0.050) + K_{1} K_{w}}{K_{1} + (0.050)}} pH = - \log {[H^{-}]}_{y} H^{-}$

02

Step 2: Calculate the ph and concentration of ions.

Calculate The ph of solution

$[{SO}_{3}^{_{2 -}}] + H_{2} O 鈼� {HSO}_{3}^{-} + {OH}^{-} \frac{x^{2}}{0.050 - x} K_{bl} = \frac{K_{w}}{K_{22}} = 1.49 脳 10^{- 7} [{HSO}_{3}^{_{-}}] = x = 8.6 脳 10^{- 5} M [H^{+}] = \frac{K_{w}}{x} = 1.16 脳 10^{- 10} M = pH = 9.94 n [{SO}_{3}^{2 -}] = 0.050 - x = 0.050 M [H_{2} {SO}_{3}] = \frac{[H^{+}] [{HSO}_{3}^{-}]}{K_{1}} = 7.2 脳 10^{- 13} M$

The concentration of solution is $= 7.2 脳 10^{- 13} M$

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Most popular questions from this chapter

(a) Calculate the quotient in $[H_{3} {PO}_{4}] / [H_{3} {PO}^{-}_{4}] in 0.0500 M {KH}_{2} {PH}_{4}$ .

(b) Find the same quotient for $0.0500 {MK}_{2} {PO}_{4}$ .

Heterogeneous equilibrium. ${CO}_{2}$ dissolves in water to give "carbonic acid" (which is mostly dissolved ${CO}_{2}$ , as described in Box 6-4).

${CO}_{2} (g) 饾啅 {CO}_{2} (aq) K = 10^{- 1.5}$

(The equilibrium constant is called the Henry's law constant for carbon dioxide, because Henry's law states that the solubility of a gas in a liquid is proportional to the pressure of the gas.) The acid dissociation constants listed for "carbonic acid" in Appendix G apply to ${CO}_{2} (aq)$ . Given that $P_{{CO}_{2}}$ in the atmosphere is $10^{- 3.4}$ atm, find the $pH$ of water in equilibrium with the atmosphere.

Calculate the pH of a $0.010 M$ solution of each amino acid in the form drawn here

Find the pH and the concentration of each species of lysine in a solution of $^{0.010}$ 0 M lysine? HCl, lysine monohydrochloride. The notation 鈥渓ysine? HCl鈥� refers to a neutral lysine molecule that has takenon one extra proton by addition of one mole of HCl. A more meaning-

ful notation shows the salt (lysineH1)( $^{C 12}$ ) formed in the reaction..

(c) How many milliliters of $0.320 {MNHO}_{3}$ should be added to $4.00 g$ of to give a pH of $10.00$ in $250 mL$ ?

See all solutions

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