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Chapter 10: Q7TY (page 223)

The both $^{[H_{2} A] = [H A^{‐}]}$ are equal as found to be $^{pH = 10.54}$

Short Answer

Expert verified

The both $^{[H_{2} A] = [H A^{‐}]}$ are found to be equal $^{pH = 10.54}$

Step by step solution

01

Definition of Henderson-Hassel Balch equation

The Henderson Hasselbalch equation approximates the link between a solution's pH or pOH, its pKa or pKb, and the ratio of dissociated chemical species concentrations.

02

Determine the both are equal as found to be

The both $^{[H_{2} A] = [H A^{‐}]}$ are equal, and the pH must be determined.

Henderson-Hassel Balch equation

$pH = {pK}_{1} + \log \frac{[H A^{-}]}{[H_{2} A]} pH = {pK}_{2} + \log \frac{[A^{2}]}{[H A^{-}]}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

It are aware of this.

The purity of the intermediate species' pH, $^{H_{2} {Arg}^{+}}$

Arginine is a kind of amino acid $^{[{Arg}^{-}]}$ at $^{{pK}_{3} = 12.1}$ is 4.00

$H_{2} {Arg}^{+} \approx \frac{1}{2} ({pK}_{1} + {pK}_{2}) = 5.40$

$^{pH = 10.54}$ (and a pH exceeding $^{= {pK}_{2}}$ )

Therefore $^{[H_{2} {Arg}^{+}] = [{Arg}^{-}]}$ at $^{pH = 10.54}$

Both of them $^{[H_{2} A] = [H A^{‐}]}$ are found to be equal $^{pH = 10.54}$ .

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Most popular questions from this chapter

Fractional composition in a triprotic system. For a triprotic system, the fractional composition equations where are $D = {[H^{+}]}^{3} + K_{1} {[H^{+}]}^{2} + K_{1} K_{2} [H^{+}] + K_{1} K_{2} K_{3}$ Use these equations to create a fractional composition diagram analogous to Figure 10-4 for the amino acid tyrosine. What is the fraction of each species at $pH 10.00$ ?

$α_{H_{3} A} = \frac{{[H^{+}]}^{3}}{D} α_{H^{2}} = \frac{K_{1} K_{2} [H^{+}]}{D} α_{H_{2} A} = \frac{K_{1} {[H^{+}]}^{2}}{D} α_{A^{3}} = \frac{K_{1} K_{2} K_{3}}{D}$

Consider the diprotic acid $H_{2} A$ with $K_{1} = 1.00 \times 10^{- 4}$ and $K_{2} = 1.00 \times 10^{- 8}$ .Find the pH and concentrations of $H_{2} A, {HA}^{-}$ , and $A^{2 -}$ in

(a)role="math" localid="1654926233413" $0.100 {MH}_{2} A;$

(b) role="math" localid="1654926403235" $0.100 MNaHA$ ;

(c) $0.100 {MNa}_{2} A$ .

Compute the pH of $^{0.010 M}$ HHis.

$C O_{2} (g) 𝆏 C O_{2} (a q) K_{H} = \frac{[C O_{2} (a q)]}{P_{C O_{2}}} = 10^{- 1.2073} m o l k g^{- 1} b a r^{- 1} a t 0^{\circ} C = 10^{- 1.6048} m o l k g^{- 1} b a r^{- 1} a t 30^{\circ} C C O_{2} (a q) + H_{2} O 𝆏 H C O_{3}^{-} + H^{+} K_{a 1} = \frac{[H C O_{3}^{-}] [H^{+}]}{[C O_{2} (a q)]} = 10^{- 6.1004} m o l k g^{- 1} a t 0^{\circ} C = 10^{- 5.8008} m o l k g^{- 1} b a r^{- 1} a t 30^{\circ} C H C O_{3}^{-} 𝆏 C O_{3}^{2 -} + H^{+} C a C O_{3} (S, a r a g o n i t e) 𝆏 C a^{2 +} + C O_{3}^{- 2} K_{s p}^{a r g} = [C a^{2 +}] [C O_{3}^{2 -}] = 10^{- 6.1113} m o l^{2} k g^{- 2} b a r^{- 2} a t 0^{\circ} C = 10^{- 6.1391} m o l^{2} k g^{- 2} b a r^{- 2} a t 30^{\circ} C C a C O_{3} (s, c a l c i t e) 𝆏 C a^{2 +} + C O_{3}^{- 2} k_{s p}^{c a l} = [C a^{2 +}] [C O_{3}^{2 -}] = 10^{- 6.3652} m o l^{2} k g^{- 2} b a r^{- 2} a t 0^{\circ} C = 10^{- 6.3713} m o l^{2} k g^{- 2} b a r^{- 2} a t 30^{\circ} C$ 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).

Phosphate at $0.01 M$ is one of the main buffers in blood plasma, whose pH is 7.45. Would phosphate be as useful if the plasma pH were 8.5?

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