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Chapter 8: Q6 TY (page 169)

Find $[H^{+}]$ and the pH of $0.05 {MI}^{2} {NO}_{3}$ .

Short Answer

Expert verified

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

The pH is 6.99.

Step by step solution

01

Concept used.

pH of a solution:

The pH of a solution can be calculated by,

$pH =- {logA}_{H^{+}}$

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

With the help of above equation, the negative logarithm of activity of $H^{+}$ ion can be calculated.

02

Step 2: Calculate the value of [H+] and the pH of .

At an ionic strength $(μ)$ of $0.050 M, γ_{H^{+}} = 0.86$ and $γ_{O H} = 0.81$ from the table 8 - 1.

role="math" localid="1663412820591" $K_{w} = [H^{+}] γ_{H^{+}} [{OH}^{-}] γ_{{OH}^{-}} = (x) (0.86) (x) (0.81) = 1.0 \times 10^{- 14} x = [H^{+}] = 1.2 \times 10^{- 7} M$

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

By using the concentration and activity coefficient of ion calculate the pH

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

The pH is 6.99.

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Most popular questions from this chapter

Write the mass balance for a saturated solution of $Ba {({HsO}_{4})}_{2}$ if the species in solution are $B a^{2 +}, b a S O_{4} (a q), H S O_{4}^{-}, S O_{4}^{2 -}$ and ${BaOH}^{+}$ .SO

Ammonia Equilibrium treated by solver. We now use the solve spreadsheet introduced in Figure 8 - 9 for ${TIN}_{3}$ solubility to find the concentration of species in 0.01 M ammonia solution, neglecting activity coefficient. In the systematic treatment of equilibrium of ${NH}_{3}$ hydrolysis, we have four unknowns $([{NH}_{3}], [{NH}_{4}^{+}], [H^{+}], [{OH}^{-}])$ and two equilibrium (8-13) , (8-14). Therefore we will estimate the concentration of 4unknowns - 2equilibirum = 2species, for which I choose localid="1663566766281" ${NH}^{+} and {OH}^{-}$ . Setup the spreadsheet shown below, in which the estimate localid="1663566820791" ${pNH}_{4}^{+} = 3 . {pOH}^{-}$ = 3 appears in B6andB7 . (Estimates comes from the $K_{b}$ equilibrium 8-17 with $[{NH}_{4}^{+}] = [{OH}^{-}] = \sqrt{K_{b} [{NH}_{3}]} \approx \sqrt{10^{- 4.755} [0.01]} \Rightarrow {pNH}_{4}^{+} = {pOH}^{-} \approx 3$ . Estimate donot have to be very good for Solver to work. The formula in cell C8 is $[{NH}_{3}] = [{NH}_{4}^{+}]$ .

$[{OH}^{-}] / K_{b}$ and the formula in the cell C9 is $[H^{+}] = K_{w} / [{OH}^{-}]$ . The mass balance $b_{1}$ appears in cell F6 and the charge balance $b_{2}$ appears in cell F7 . Cell F8 has the sum $b_{1}^{2} + b_{2}^{2}$ . As described for ${TIN}_{3}$ on page 176, open the solver window and set the Solver Option. Then use the Solver to set the target cell F8 Equal to Min by changing cells B6 : B7 . What are the concentrations of the species? What fraction of ammonia $(= [{NH}_{4}^{+}] / [{NH}_{4}^{+}] + [{NH}_{3}])$ is hydrolyzed. Your answer should agree with those from Goal Seek in Figure 8-8

Which statements are true In the ionic strength range $0_{I} 0.1 M,$

activity coefficients decrease with

a) increasing ionic strength

b) increasing ionic charge

c) decreasing hydrated radius

(a) Following the example of Mg(OH)₂ in Section 8-5, write the equations needed to find the solubility of Ca(OH)₂. Include activity coefficients where appropriate. Look up the equilibrium constants in Appendixes F and I.

(b) Suppose that the size of CaOH⁺= Ca(H₂O)₅(OH)⁺ is 500 pm. Including activity coefficients, compute the concentrations of all species, the fraction of hydrolysis (= [CaOH⁺]/{[Ca²⁺] + [CaOH⁺]}), and the solubility of Ca(OH)₂ in g/L. The Handbook of Chemistry and Physics lists the solubility of Ca(OH)₂ as 1.85 g/L at 0⁰C and 0.77 g/L at 100⁰C

Write the charge balance for a solution containing $H^{+}, {OH}^{-}, {Ca}^{2 +}, {HCO}_{3}^{-}, {CO}_{3}^{2 -}, Ca {({HCO}_{3})}^{+}, Ca {(OH)}^{+}, K^{+} and {ClO}_{4}^{-}$

See all solutions

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