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Chapter 8: QDE (page 182)

Using activities, calculate the pH and concentration of $H^{+}$ in 0.050MLiBr at $25 ° C$ .

Short Answer

Expert verified

The pH of the given solution is 6.99

Step by step solution

01

Define the formula for PH calculation.

The pH of a solution can be calculated as,

$p H = - l o g A_{H^{+}} = - l o g [H^{+}] γ_{H^{+}}$

where,

$A_{H^{+}}$ - activity of hydrogne ion

$γ_{H^{+}}$ - activity coefficient of hydrogne ion.

02

Calculate the PH form the activity.

Given

0.050MLiBr at $25 ° C$

At an ionic strength role="math" localid="1654765585845" $(μ)$ of and $γ_{OH} = 0.81$ .

role="math" localid="1654765776290" $[H^{+}] γ_{H^{+}} [{OH}^{-}] γ_{OH} = (x) (0.86) (x) (0.81) = 1.0 \times 10^{- 14} x = 1.2 \times 10^{- 7} M .$

The concentration of $H^{+}$ ion is $1.2 \times 10^{- 7} M$

Calculate the pH by using the concentration of $H^{+}$ ion

$pH = - \log [(1.2 \times 10^{- 7}) (0.86)] = 6.99$

Thus, theof the given solution is 6.99 .

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Most popular questions from this chapter

Systematic treatment of equilibrium for ion pairing. Let’s derive the fraction of ion pairing for the salt in Box $8 - 1$ , which are $0.025 F NaCI, {Na}_{2} {SO}_{4}, {MgCI}_{2}, {MgSO}_{4}$ . Each case is somewhat different. All of the solutions will be near neutral pH because hydrolysis reactions of ${Mg}^{2 +}, {SO}^{2 -}_{4}, {Na}^{+}, {CI}^{-}$ have small equilibrium constants. Therefore, we assume that $[H^{+}] = [{OH}^{-}]$ and omit these species from the calculations. We work ${MgCI}_{2}$ as an example and then you asked to work each of the others. The ion-pair equilibrium constant, $K_{ip}$ comes from Appendix J.

Pertinent reaction:

${Mg}^{2 +} {CI}^{-} ֏ {MgCI}^{+} (aq) K_{ip} = \frac{[{MgCI}^{+} (aq)] γ_{{MgCI}^{+}}}{[{Mg}^{2 +}] γ_{{Mg}^{2 +}} [{CI}^{-}] γ_{{CI}^{-}}} {logK}_{ip} = 0.6 . {pK}_{ip} = - 0.6 \to (A)$

Charge balance (omitting $H^{+}, {OH}^{-}$ whose concentrations are both small in comparison with $[{Mg}^{+}], [{MgCI}^{+}], [{CI}^{-}]$ :

role="math" localid="1655088043259" $2 [{Mg}^{2 -}] + [{MgCI}^{+}] = [{CI}^{-}] \to (B)$

Mass balance:

$[{Mg}^{2 -}] + [{MgCI}^{+}] = F = 0.025 M \to (C) [{CI}^{-}] + [{MgCI}^{+}] = 2 F = 0.050 M \to (D)$

Only two of the three equations (B),(C) and (D) are independent. If you double (c) and subtract (D) , you will produce (B). we choose (C) and (D) as independent equations.

Equilibrium constant expression : Equation (A)

Count : 3 equations (A,C,D) and 3 unknowns $([{Mg}^{2 +}], [{MgCI}^{+}], [{CI}^{-}])$

Solve: We will use Solver to find

$(number of unknowns) - (number of equiliberia) = 3 - 1 = 2$ unknown concentrations.

The spreadsheet shows the work. Formal concentration $F = 0.0025 M$ appears in cell $G 2$ . We estimate ${pMg}^{2 +}, {pCI}^{-}$ in cell $B 8$ and $B 9$ . The ionic strength in cell $B 5$ is given by the formula in cell $H 24$ . Excel must be set to allow for circular definitions as described on page role="math" localid="1655088766279" $179$ . The sizes of role="math" localid="1655088853561" ${Mg}^{2 +}, {CI}^{-}$ are from Table $8 - 1$ and the size of ${MgCI}^{+}$ is a guess. Activity coefficient are computed in columns $E, F$ . Mass balance $b_{1} = F - [{Mg}^{2 +}] - [{MGCI}^{+}], b_{2} = 2 F - [{CI}^{-}] - [{MgCI}^{+}]$ appears in cell $H 14, H 15$ , and the sum of squares ${b^{2}}_{1} + {b^{2}}_{2}$ appears in cell $H 16$ . The charge balance is not used because it is not independentof the two mass balances.

Solver is invoked to minimizes ${b^{2}}_{1} + {b^{2}}_{2}$ in cell $H 16$ be varying ${pMg}^{2 +}, {pCI}^{-}$ in cells $B 8$ and $B 9$ . From the optimized concentration, the ion-pair fraction $= \frac{[{MgCI}^{+}]}{F} = 0.0815$ is computed in cell $D 15$ .

The problem: Create a spreadsheet like the one for ${MgCI}^{+}$ to find the concentration, ionic strength, and ion pair fraction in $0.025 M NaCI$ . The ion pair formation constant from Appendix J is log $K_{ip} = 10^{- 0.5}$ for the reaction ${Na}^{+} + {CI}^{-} ֏ NaCI (aq)$ . The two mass balances are $[{Na}^{+}] + [NaCI (aq)] = F, [{Na}^{+}] = [{CI}^{-}]$ Estimate ${pNa}^{+}, {pCI}^{-}$ for input and then minimizes the sum of square of the two mass balances.

Interpolate in Table 8-1 to find the activity coefficient of $H^{+}$ when $μ = 0.030 M$

Write the charge balance for an aqueous solution of arsenic acid, $H_{3} {AsO}_{4}$ , in which the acid can disassociate to role="math" localid="1654936423245" $H_{3} {AsO}^{-}_{4}, H {AsO}_{4}^{2 -}, {andAsO}_{4}^{3 -}$ . Look up the structure of arsenic acid in Appendix G and write the structure of ${HAsO}^{2 -}_{4}$

(a) Write the mass balance for ${CaCl}_{2}$ in water if the species are ${Ca}^{2 +}$ and ${CI}^{-}$ .

(b) Write the mass balance if the species are ${Ca}^{2 +}, {CI}^{-}, {CaCI}^{+}$ , and ${CaOH}^{+}$

(c) Write the charge balance for part (b).

Color Plate 4 shows how the color of the acid-base indicatorbromocresol green (H2BG) changes as NaCl is added to an aqueoussolution of (H1)(HBG2). Explain why the color changes from palegreen to pale blue as NaCl is added.

See all solutions

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