60.0 mL of 0.45 M #K_3PO_4# is mixed with 240.0 mL of 0.20 M #K_2SO_3#. What is the final concentration of all three ions in the solution?

Question

Caleb Adams · Accepted Answer

#["K"^(+)] = "0.59 M"" "# #["PO"_4^(3-)] = "0.090 M"" "# #["SO"_3^(2-)] = "0.16 M"#

The dissociation of the two salts—which are both soluble in aqueous solution—to produce negatively charged anions and positively charged cations is the key to solving this problem.

Tripotassium phosphate, #"K"_3"PO"_4#, will dissociate in aqueous solution to form

#"K"_3"PO"_text(4(aq]) -> color(red)(3)"K"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#

Potassium sulfite, #"K"_2"SO"_3#, will dissociate in aqueous solution to form

#"K"_2"SO"_text(3(aq]) -> color(blue)(2)"K"_text((aq])^(+) + "SO"_text(3(aq])^(2-)#

This indicates that the target solution's three ions will be

Two pieces of information are necessary in order to calculate the concentration of each ion in the finished solution.

Given the volume and molarities of the two solutions, you can calculate the number of moles of each salt you're mixing by using the definition of molarity, which is the number of moles of solute per liter of solution.

#color(blue)(c = n/V implies n = c * V)#

#n_(K_3PO_4) = "0.45 M" * 60.0 * 10^(-3)"L" = "0.0270 moles K"_3"PO"_4#

additionally

#n_(K_2SO_3) = "0.20 M" * 240.0 * 10^(-3)"L" = "0.0480 moles K"_2"SO"_3#

Find out how many moles of each ion are created in solution when the salts dissolve by examining the two balanced chemical equations.

Tripotassium phosphate will produce #color(red)(3)# moles of potassium cations and #1# mole of phosphate anions, so you can say that

#n_(K^(+)) = color(red)(3) xx "0.0270 moles" = "0.0810 moles K"^(+)#

#n_(PO_4^(3-)) = 1 xx "0.0270 moles" = "0.0270 moles PO"_4^(3-)#

Potassium sulfite will produce #color(blue)(2)# moles of potassium cations and #1# mole of sulfite anions, so you can say that

#n_(K^(+)) = color(blue)(2) xx "0.0480 moles" = "0.0960 moles K"^(+)#

#n_(SO_3^(2-)) = 1 xx "0.0480 moles" = "0.0480 moles SO"_3^(2-)#

As you can see, both salts are contributing potassium cations to the final solution, so you will have

#n_(K^(+)"total") = "0.0810 moles" + "0.0960 moles" = "0.177 moles K"^(+)#

The finished solution's overall volume will be

#n_"total" = V_(K_3PO_4) + V_(K_2SO_3)#

#n_"total" = "60.0 mL" + "240.0 mL" = "300.0 mL"#

Consequently, in the finished solution, each ion's concentration will be

#["K"^(+)] = "0.177 moles"/(300.0 * 10^(-3)"L") = color(green)("0.59 M")#

#["PO"_4^(3-)] = "0.0270 moles"/(300.0 * 10^(-3)"L") = color(green)("0.090 M")#

#["SO"_3^(2-)] = "0.0480 moles"/(300.0 * 10^(-3)"L") = color(green)("0.16 M")#

Two sig figs are used to round the answers.

Ellie Andrews · Answer

undefined The final concentration of all three ions in the solution would be as follows:

- For K⁺ ions: $0.45 \, \text{M} \times \left(\frac{60.0}{300.0}\right) + 0.20 \, \text{M} \times \left(\frac{240.0}{300.0}\right)$
- For PO₄³⁻ ions: $0.45 \, \text{M} \times \left(\frac{60.0}{300.0}\right)$
- For SO₃²⁻ ions: $0.20 \, \text{M} \times \left(\frac{240.0}{300.0}\right)$

Xavier Avery · Answer

undefined To find the final concentration of all three ions in the solution, you need to first calculate the moles of each ion present in the initial solutions, then add them up to find the total moles of each ion in the final solution. Finally, divide the total moles by the total volume of the final solution to find the final concentration of each ion.

For K3PO4:
Initial moles = volume (L) × concentration (mol/L)
Initial moles of K3PO4 = 0.060 L × 0.45 mol/L = 0.027 moles

For K2SO3:
Initial moles = volume (L) × concentration (mol/L)
Initial moles of K2SO3 = 0.240 L × 0.20 mol/L = 0.048 moles

Total moles of K ions in the final solution:
Total moles of K ions = moles of K3PO4 + moles of K2SO3
Total moles of K ions = 0.027 moles + 2 × 0.048 moles = 0.123 moles

Total moles of P ions in the final solution:
Total moles of P ions = moles of K3PO4
Total moles of P ions = 0.027 moles

Total moles of SO3 ions in the final solution:
Total moles of SO3 ions = moles of K2SO3
Total moles of SO3 ions = 2 × 0.048 moles = 0.096 moles

Final concentration of K ions:
Final concentration of K ions = Total moles of K ions / Total volume of solution
Final concentration of K ions = 0.123 moles / (0.060 L + 0.240 L) = 0.307 M

Final concentration of P ions:
Final concentration of P ions = Total moles of P ions / Total volume of solution
Final concentration of P ions = 0.027 moles / (0.060 L + 0.240 L) = 0.0675 M

Final concentration of SO3 ions:
Final concentration of SO3 ions = Total moles of SO3 ions / Total volume of solution
Final concentration of SO3 ions = 0.096 moles / (0.060 L + 0.240 L) = 0.24 M