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Chapter 16: Problem 52

A solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{BaCl}_{2}\) and \(125 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\). What are the concentrations of barium and sulfate ions in this solution? Assume only \(\mathrm{SO}_{4}^{2-}\) ions (no \(\left.\mathrm{HSO}_{4}^{-}\right)\) are present.

Short Answer

Expert verified
To find the final concentrations of barium and sulfate ions, we first calculate the moles of each ion in their respective solutions: moles(Ba^2+) \(= (0.020 \, M)(75.0 \, mL)(1 \, L / 1000 \, mL)\) moles(SO鈧劼测伝) \(= (0.040 \, M)(125 \, mL)(1 \, L / 1000 \, mL)\) Next, we find the total moles of each ion, the total volume of the mixed solution, and finally divide the total moles of each ion by the total volume to find their final concentrations: Final concentration(Ba^2+) \(= \frac{\text{Total moles(Ba}^{2+}\text{)}}{\text{Total volume}}\) Final concentration(SO鈧劼测伝) \(= \frac{\text{Total moles(SO}鈧刕{2-}\text{)}}{\text{Total volume}}\)

Step by step solution

01

Find moles of ions in each solution

First, we need to find the moles of Ba^2+ and SO鈧劼测伝 in each solution using the given concentrations and volumes. To do this, we will use the formula: moles = concentration 脳 volume For Ba^2+ from BaCl鈧�: moles(Ba^2+) = (0.020 M) 脳 (75.0 mL) 脳 (1 L / 1000 mL) For SO鈧劼测伝 from K鈧係O鈧�: moles(SO鈧劼测伝) = (0.040 M) 脳 (125 mL) 脳 (1 L / 1000 mL)
02

Calculate the total moles of each ion

Now, we will add the moles of Ba^2+ and SO鈧劼测伝 found in each solution to get their total moles. Total moles(Ba^2+) = moles(Ba^2+) from BaCl鈧� Total moles(SO鈧劼测伝) = moles(SO鈧劼测伝) from K鈧係O鈧�
03

Calculate the total volume of the mixed solution

We need to find the total volume of the mixed solution to calculate the final concentrations of the ions. To do this, simply add the individual volumes: Total volume = Volume of BaCl鈧� solution + Volume of K鈧係O鈧� solution
04

Calculate the final concentrations of ions

Finally, we will divide the total moles of each ion by the total volume of the mixed solution to find their final concentrations. Final concentration(Ba^2+) = Total moles(Ba^2+) / Total volume Final concentration(SO鈧劼测伝) = Total moles(SO鈧劼测伝) / Total volume

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Most popular questions from this chapter

Solutions of sodium thiosulfate are used to dissolve unexposed \(\mathrm{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for blackand-white film. What mass of AgBr can dissolve in \(1.00 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\) reacts with \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\) to form a complex ion:

Explain the following phenomenon: You have a test tube with about \(20 \mathrm{~mL}\) silver nitrate solution. Upon adding a few drops of sodium chromate solution, you notice a red solid forming in a relatively clear solution. Upon adding a few drops of a sodium chloride solution to the same test tube, you notice \(a\) white solid and a pale yellow solution. Use the \(K_{s p}\) values in the book to support your explanation, and include the balanced equations.

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a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\), calculate the value for the equilibrium constant for the following reaction: \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 \mathrm{M} .\)
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