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Chapter 10: Q2TY (page 219)

Find the pH with $1.50 {gNaN}_{2} P$ instead of 1.20 g.

Short Answer

Expert verified

The pH of given ${Na}_{2} P$ solution is 5.57.

Step by step solution

01

Concept used.

Henderson-Hassel Balch equation

$pH = {pK}_{1} + \log \frac{| {IA}^{-} |}{| H_{2} A |} pH = {pK}_{1} + \log \frac{| A^{2} |}{| {HA}^{-} |}$

02

Step2: Calculate the pHwith 1.50NaN2P instead of 1.20 g.

Disodium phthalate $C_{8} H_{4} O_{4} {Na}_{2} = 210 g / mol$

$pH = {pK}_{2} + \log \frac{[A^{2}]}{|{HA}^{2}|} pH = {pK}_{2} + \log \frac{[p^{2}]}{|HP -|} = 5.408 + \log \frac{(1.50 g) / (210.094 g / mol)}{(1 . 00_{g}) (204.22 g_{g / mol)}} = 5.57$

The pH of Na₂P solution as found to be 5.57.

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Most popular questions from this chapter

Fractional composition in a tetraprotic system. Prepare a fractional composition diagram analogous to Figure 10-4 for the tetraprotic system derived from hydrolysis of $^{{Cr}^{+}}$ :

localid="1654853037629" $C r^{3 +} + H_{2} O ⇋ C r {(O H)}^{2 +} + H^{+} K_{a 1} = 10^{- 3.80} C r {(O H)}^{2 +} + H_{2} O ⇋ C r {(O H)}_{2}^{+} + H^{+} K_{a 2} = 10^{- 6.40} C r {(O H)}_{2}^{+} + H_{2} O ⇋ C r {(O H)}_{3} (a q) + H^{+} K_{a 3} = 10^{- 6.40} C r {(O H)}_{3} (a q) + H_{2} O ⇋ C r {(O H)}_{4}^{-} + H^{+} K_{a 4} =^{- 11.40}$

(Yes, the values oflocalid="1654853051658" $^{K_{a 2}}$ andlocalid="1654853058271" $^{K_{a 3}}$ are equal.)

(a) Use these equilibrium constants to prepare a fractional composition diagram for this tetraprotic system.

(b) You should do this part with your head and your calculator, not your spreadsheet. The solubility of is given by

localid="1654853063951" $C r {(O H)}_{3} (S) C r {(O H)}_{3} (a q) K_{a 3} = 10^{- 6.80}$

What concentration of localid="1654853075036" $^{Cr {(OH)}_{3} (aq)}$ is in equilibrium with solid localid="1654853085944" $^{C r {(O H)}_{3} (a q) (S)}$ ?

(c) What concentration oflocalid="1654853094735" $^{C r {(O H)}^{2 +}}$ is in equilibrium with localid="1654853101499" $^{C r {(O H)}_{3} (a q) (S)}$ if the solution localid="1654853109266" $^{p H}$ is adjusted tolocalid="1654853117749" $^{4.00}$ ?

$C O_{2} (g) 𝆏 C O_{2} (a q) K_{H} = \frac{[C O_{2} (a q)]}{P_{C O_{2}}} = 10^{- 1.2073} m o l k g^{- 1} b a r^{- 1} a t 0^{\circ} C = 10^{- 1.6048} m o l k g^{- 1} b a r^{- 1} a t 30^{\circ} C C O_{2} (a q) + H_{2} O 𝆏 H C O_{3}^{-} + H^{+} K_{a 1} = \frac{[H C O_{3}^{-}] [H^{+}]}{[C O_{2} (a q)]} = 10^{- 6.1004} m o l k g^{- 1} a t 0^{\circ} C = 10^{- 5.8008} m o l k g^{- 1} b a r^{- 1} a t 30^{\circ} C H C O_{3}^{-} 𝆏 C O_{3}^{2 -} + H^{+} C a C O_{3} (S, a r a g o n i t e) 𝆏 C a^{2 +} + C O_{3}^{- 2} K_{s p}^{a r g} = [C a^{2 +}] [C O_{3}^{2 -}] = 10^{- 6.1113} m o l^{2} k g^{- 2} b a r^{- 2} a t 0^{\circ} C = 10^{- 6.1391} m o l^{2} k g^{- 2} b a r^{- 2} a t 30^{\circ} C C a C O_{3} (s, c a l c i t e) 𝆏 C a^{2 +} + C O_{3}^{- 2} k_{s p}^{c a l} = [C a^{2 +}] [C O_{3}^{2 -}] = 10^{- 6.3652} m o l^{2} k g^{- 2} b a r^{- 2} a t 0^{\circ} C = 10^{- 6.3713} m o l^{2} k g^{- 2} b a r^{- 2} a t 30^{\circ} C$ Effect of temperature on carbonic acid acidity and the solubility of localid="1654949830957" ${CaCO}_{3} \times^{14} Box 10 - 1$ states that marine life withlocalid="1654949841354" ${CaCO}_{3}$ shells and skeletons will be threatened with extinction in cold polar waters

before that will happen in warm tropical waters. The following equilibrium constants apply to seawater at $0^{\circ}$ ndlocalid="1654949866125" $30^{\circ} C$ , when concentrations are measured in moles per kilogram of seawater and pressure is in bars:

localid="1654951136007" ${CO}_{2} (g) 𝆏 {CO}_{2} (aq) K_{H} = \frac{[{CO}_{2} (aq)]}{P_{{CO}_{2}}} = 10^{- 1.2073} mol {kg}^{- 1} {bar}^{- 1} at 0^{\circ} C = 10^{- 1.6048} mol {kg}^{- 1} {bar}^{- 1} at 30^{\circ} C {CO}_{2} (aq) + H_{2} O 𝆏 {HCO}_{3}^{-} + H^{+} K_{a 1} = \frac{[{HCO}_{3}^{-}] [H^{+}]}{[{CO}_{2} (aq)]} = 10^{- 6.1004} mol {kg}^{- 1} at 0^{\circ} C = 10^{- 5.8008} mol {kg}^{- 1} {bar}^{- 1} at 30^{\circ} C {HCO}_{3}^{-} 𝆏 {CO}_{3}^{2 -} + H^{+} {CaCO}_{3} (S, aragonite) 𝆏 {Ca}^{2 +} + {CO}_{3}^{- 2} K_{sp}^{\arg} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = 10^{- 6.1113} {mol}^{2} {kg}^{- 2} {bar}^{- 2} at 0^{\circ} C = 10^{- 6.1391} {mol}^{2} {kg}^{- 2} {bar}^{- 2} at 30^{\circ} C {CaCO}_{3} (s, calcite) 𝆏 {Ca}^{2 +} + {CO}_{3}^{- 2} k_{sp}^{cal} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = 10^{- 6.3652} {mol}^{2} {kg}^{- 2} {bar}^{- 2} at 0^{\circ} C = 10^{- 6.3713} {mol}^{2} {kg}^{- 2} {bar}^{- 2} at 30^{\circ} C$

The first equilibrium constant is calledlocalid="1654949908801" $K_{H}$ for Henry's law (Problem 10-10). Units are given to remind you what units you must use.

(a) Combine the expressions forlocalid="1654949922835" $K_{H}, K_{21}$ , andlocalid="1654949935065" $K_{22}$ to find an expression forlocalid="1654949944862" $[{CO}_{3}^{- 2}]$ in terms oflocalid="1654949958224" $P_{{CO}_{2}}$ andlocalid="1654949971486" $[H^{+}]$ .

(b) From the result of (a), calculatelocalid="1654949980228" $[{CO}_{3}^{2 -}] ({molkg}^{- 1})$ atlocalid="1654950000228" $p_{{CO}_{2}} = 800 μ$ bar and pH=7.8 at temperatures oflocalid="1654950013597" $0^{\circ}$ (polar ocean) andlocalid="1654950030147" $30^{\circ} C$ (tropical ocean). These are conditions that could possibly be reached around the year 2100 .

(c) The concentration oflocalid="1654950042348" ${Ca}^{2 +}$ in the ocean islocalid="1654950053646" $0.010 M$ . Predict whether aragonite and calcite will dissolve under the conditions in (b).

The diprotic acid $H_{2} A$ has ${pK}_{1} = 4.00$ and ${pK}_{2} = 8.00$ .

(a) At what pH is $[H_{2} A]$ = $[H A^{-}]$ ?

(b) At what pH is $[H A^{-}]$ = $[A^{2 -}]$ ?

(c) Which is the principal species at pH 2.00: $H_{2} A, {HA}^{-}$ or $A^{2 -}$ ?

(d) Which is the principal species at pH 6.00?

(e) Which is the principal species at pH 10.00?

(a) Using activity coefficients, calculate the pH of a solution containing a 2.00:1.00 mole ratio of ${HC}^{2} : C^{3 -}$ ( $H_{3} C$ = citric acid). The ionic strength is 0.010 M.

(b) What will be the pH if the ionic strength is raised to 0.10 M and the mole ratio ${HC}^{- 2} : C^{3 -}$ is kept constant?

Draw the structure of the predominant form of pyridoxal-

5-phosphate at pH 7.00

See all solutions

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