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Chapter 16: Problem 46

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in 500.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 10.0 \mathrm{~mL}\) of $0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\( diluted to \)500.0 \mathrm{~mL},(\mathbf{d})\( a solution formed by mixing \)20.0 \mathrm{~mL}$ of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of $8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}$.

Short Answer

Expert verified
For the given strong base solutions, we find the following: a) 0.182 M KOH: \([\mathrm{OH}^{-}] = 0.182 \mathrm{M}\) and \(pH \approx 13.26\) b) 3.165 g of KOH in 500.0 mL solution: \([\mathrm{OH}^{-}] = 0.1128 \mathrm{M}\) and \(pH \approx 13.05\) c) 10.0 mL of 0.0105 M Ca(OH)₂ diluted to 500.0 mL: \([\mathrm{OH}^{-}] = 0.00042 \mathrm{M}\) and \(pH \approx 10.62\) d) Solution formed by mixing 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 × 10⁻³ M NaOH: \([\mathrm{OH}^{-}] = 0.01547 \mathrm{M}\) and \(pH \approx 12.19\)

Step by step solution

01

Find \([\mathrm{OH}^{-}]\)

Since KOH is a strong base, it will completely dissociate in water: \[ \mathrm{KOH} \rightarrow \mathrm{K}^{+} + \mathrm{OH}^{-}\] Therefore, \([\mathrm{OH}^{-}] = 0.182 \mathrm{M}\).
02

Calculate the pOH

Now, we need to find the pOH of the solution using the formula: \[pOH = -\log{[\mathrm{OH}^{-}]}\] Substituting the values, we get: \(pOH = -\log{(0.182)} \approx 0.74\)
03

Determine the pH

To find the pH, we use the relationship between pH and pOH: \(pH = 14 - pOH\) \(pH = 14 - 0.74 \approx 13.26\) So, for this solution, \([\mathrm{OH}^{-}] = 0.182 \mathrm{M}\) and \(pH \approx 13.26\). #b) Calculate \([\mathrm{OH}^{-}]\) and pH for a solution formed with 3.165 g of KOH in 500.0 mL#
04

Calculate the concentration of KOH

We need to find the molar concentration of the solution. The molar mass of KOH is \((39.10 + 16.00 + 1.01) = 56.11 \ g/mol\). First, we calculate moles of KOH dissolved: \(\frac{3.165g}{56.11g/mol} \approx 0.0564\ mol\) Now, we find the concentration of KOH: \(\frac{0.0564\ mol}{0.5\ L} = 0.1128\ M\)
05

Calculate \([\mathrm{OH}^{-}]\)

Just like in part a, \([\mathrm{OH}^{-}] = 0.1128\ M\).
06

Determine the pOH and pH

Using our formula, we find pOH: \(pOH = -\log{(0.1128)} \approx 0.95\) And finally, the pH: \(pH = 14 - 0.95 \approx 13.05\) For this solution, \([\mathrm{OH}^{-}] = 0.1128 \mathrm{M}\) and \(pH \approx 13.05\). #c) Calculate \([\mathrm{OH}^{-}]\) and pH for 10.0 mL of 0.0105 M \(\mathrm{Ca}(\mathrm{OH})_{2}\) diluted to 500.0 mL#
07

Determine the new concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Using the dilution formula \(C_1V_1 = C_2V_2\), we calculate the new concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) after dilution: \(C_2 = \frac{C_1V_1}{V_2} = \frac{0.0105M \times 0.010L}{0.5L} = 0.00021\ M\)
08

Calculate \([\mathrm{OH}^{-}]\)

When \(\mathrm{Ca}(\mathrm{OH})_{2}\) dissociates, it produces 2 moles of OH⁻ ions for each mole of the compound: \[ \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\] Thus, \([\mathrm{OH}^{-}] = 2 \times 0.00021\ M = 0.00042\ M\).
09

Determine the pOH and pH

Calculating pOH and pH: \(pOH = -\log{(0.00042)} \approx 3.38\) \(pH = 14 - 3.38 \approx 10.62\) For this solution, \([\mathrm{OH}^{-}] = 0.00042 \mathrm{M}\) and \(pH \approx 10.62\). #d) Calculate \([\mathrm{OH}^{-}]\) and pH for a solution formed by mixing 20.0 mL of 0.015 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) with 40.0 mL of \(8.2 \times 10^{-3}\ \mathrm{M}\ \mathrm{NaOH}\)#
10

Find moles of OH⁻ ions from each source

For \(\mathrm{Ba}(\mathrm{OH})_{2}\): \(0.015M \times 0.020L = 3.0\times 10^{-4}\ mol\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) Since each mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\) produces 2 moles of OH⁻, we obtain \(2 \times 3.0\times 10^{-4}\ mol = 6.0\times 10^{-4} \ mol\) of \(\mathrm{OH}^{-}\). For \(\mathrm{NaOH}\): \(8.2\times 10^{-3}\ M \times 0.040L = 3.28 \times 10^{-4}\ mol\) of \(\mathrm{OH}^{-}\)
11

Calculate the total \([\mathrm{OH}^{-}]\)

Add up the moles of OH⁻ ions: \(total\ moles\ of\ OH^- = 6.0\times 10^{-4} + 3.28 \times 10^{-4} = 9.28 \times 10^{-4} \ mol\) Calculate the concentration of \(\mathrm{OH}^{-}\) knowing that the total volume of the solution is 60 mL: \([\mathrm{OH}^{-}] = \frac{9.28 \times 10^{-4}\ mol}{0.060L} = 0.01547\ M\)
12

Determine the pOH and pH

Now, calculate the pOH and pH: \(pOH = -\log{(0.01547)} \approx 1.81\) \(pH = 14 - 1.81 \approx 12.19\) In this solution, \([\mathrm{OH}^{-}] = 0.01547 \mathrm{M}\) and \(pH \approx 12.19\).

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Most popular questions from this chapter

Predict the stronger acid in each pair: (a) \(\mathrm{HCl}\) or HF; (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) or \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) or \(\mathrm{HBrO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) or \(\mathrm{HC}_{2} \mathrm{O}_{4} \overline{;} ;(\mathbf{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) or phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right) .\)

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) propylamine, $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ;$ (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-} ;(\mathbf{c})\) benzoate ion, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}$

A \(0.25 M\) solution of a salt NaA has \(\mathrm{pH}=9.29 .\) What is the value of \(K_{a}\) for the parent acid HA?

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$ \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ A \(0.035 \mathrm{M}\) solution of ephedrine has a \(\mathrm{pH}\) of 11.33 . (a) What are the equilibrium concentrations of $\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+},\( and \)\mathrm{OH}^{-} ?$ (b) Calculate \(K_{b}\) for ephedrine.

(a) Give the conjugate base of the following Brønsted Lowry acids: (i) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) (ii) HBr. (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{CN}^{-},\) (ii) \(\mathrm{HSO}_{4}^{-}\).
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