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Chapter 10: Q10DE (page 229)

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

Short Answer

Expert verified

The isoionic Ph is $= 10.37$

Step by step solution

01

Step 1: Define the term isoionic ph.

When a pure, neutral polyprotic acid is dissolved in water, the isoionic point or isoionic pH is obtained.

The isoelectric point is the pH at which the polyprotic acid's average charge is zero (0).

02

Calculate the isoionic PH

The pH of a neutral, polyprotic molecule in pure solution.

The only ions present are $H^{+}, {OH}^{-}$

Isoionic is the name given to polyprotic species.

$[H +] = \sqrt{\frac{K_{2} K_{3} F + K_{2} K_{w}}{K_{2} + F}}$

Where F is the formal concentration.

Here Arginine has four forms $H_{3} A^{2 +}, H_{2} A^{+}, HA$ and $A^{-}$

HA form shown as

$[H +] = \sqrt{\frac{K_{2} K_{3} (0.010) + K_{2} K_{w}}{K_{2} + 0.010}}$

$= 4.28 \times 10^{- 11} MpH = 10.37$

Thus the isoionic Ph is $= 10.37$

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Most popular questions from this chapter

How many millilitres of 1.00 M KOH should be added to 100 mL of solution containing 10.0 g of histidine hydrochloride

[His? HCI (HisH1) (CI2), FM 191.62 ] to get a pH of 9.30 ?

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" $^{4.00}$ ?

(a) Copy column B of your spreadsheet and paste it into column $G$ . Change $K_{1}$ to $10^{- 4}$ in cell G5 and change K₂ to 10^-8 in cell G6. Change F to 0.01 in cell G8. Column G now contains concentrations for the amphiprotic salt ${N_{a}}^{+} {HA}^{-}$ with $K_{1} = 10^{- 4}, K_{2} = 10^{- 8}$ , and $F = 0.01 M$ . Check the answers by hand beginning with $pH = \frac{1}{2} ({pK}_{1} + {pK}_{2})$ . With $[{HA}^{-}]$ $, F$ , calculate $[H_{2} A]$ and $[A^{2 -}]$ . Then find $[{HA}^{-}] > > F - [H_{2} A] - [A^{2 -}]$ .

(b) Copy column G of your spreadsheet and paste it into column H. Change K₂ to 10^-5 in cell G6. Column G now contains the concentrations for the intermediate form of a diprotic acid with K₁=10^-4, K₂₌10^-5, and F $= 0.01 M$ . You should observe $[{HA}^{-}] = 6.13 \times 10^{- 3} M$ and $pH = 4.50 .$

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

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

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

(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.

See all solutions

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