Calculate the potassium ion conc. (in mol/L) in a sol. prepared by adding 124.7 mL of a 0.412 M KNO3 (aq) sol.

Question

Calculate the potassium ion conc. (in mol/L) in a sol. prepared by adding 124.7 mL of a 0.412 M KNO3 (aq) sol. to 156.1 mL of a 0.269 M K2SO4(aq) sol. Assume volumes are additive.

If someone could show me how they get the answer to this that would be great.

Many thanks!

Answer 1

You need to figure out how many moles of potassium ion are coming from each solution, add them together, then divide by the total volume, which is 280.8 ml.

From KNO3, you have 0.1247 L * 0.412 M or 0.0514 moles

From K2SO4, you have 2 * 0.1561 L * 0.269 M or 0.0840 moles (remember, 2 K's per molecule)

Total number of moles is 0.1354 in 0.2808 L for a concentration of 0.482 M.

Answer 2

The simple maths is

(124.7 x 0.412 x 1 + 156.1 x 0.269 x 2) / (124.7 + 156.1)

Therfore [K+] = 0.482 M

This type of question always takes the form

(V1M1n1 + V2M2n2)/(V1 + V2)

where
V is volume
M Molarity
n the number of ions in the substance. For KNO3 n1 =1 and for K2SO4 n2=2

Answer 3

consistent with dilution legislations of answer,MiVi=MfVf the placement,Mi=conc. Of preliminary answer=11.6 M Vi=vol. Of preliminary answer=15 ml Mf=conc. of only spectacular answer =? Vf=vol. Of inal answer=500 ml Now,15 x 11.6=500 x Mf Mf=0.348 M (respond)