You have 95.0 mL of a 2.50 M solution of Na2CrO4(aq). You also have 125 mL of a 2.50 M solution of AgNO3(aq). Calculate the concentration of Na+ after the two solutions are mixed together.
1.08 M
0.475 M
5.00 M
0.00 M
2.16 M
mols Na from Na2CrO4 = 0.095 x 2.50 = ?
mols Na from AgNO3 = 0
total volume = (95 mL + 125 mL)/1000) = ?L
(Na) = mols Na/total volume in L.
mols Na from Na2CrO4 = 0.095 x 2.50 = .2375
mols Na from AgNO3 = 0
total volume = (95 mL + 125 mL)/1000) = .22L
(Na) = mols Na/total volume in L. .2375/.22L= 8.1.07954M
mols Na from Na2CrO4 = 0.095 x 2.50 = .2375
mols Na from AgNO3 = 0
total volume = (95 mL + 125 mL)/1000) = .22L
(Na) = mols Na/total volume in L. .2375/.22L= 1.07954M
Dr.Bob, 1.07954M or 1.08M isn't the answer but 2.16M is. so, why is that. You make a mistake there Dr.Bob222. Can you solve the problem again?
1st: 95mL÷1000mL=.090L because there are 1000mL in liter
2nd: .095mL*2.50M=.2375M per L
3rd: .2375M per Liter *2 =.475M per L because we have two Na(Na2)
4th: 95mL+125mL= 220mL
5th: 220mL÷1000=.220L
6th: .475M per L÷.220L=2.159M
Answe: Your answeris e which is 2.16M
To calculate the concentration of Na+ after the two solutions are mixed together, you need to determine the number of moles of Na+ ions present in each solution and then combine them.
First, let's find the number of moles of Na+ in the Na2CrO4 solution. The formula for calculating the number of moles is:
moles = concentration (in M) x volume (in L)
Given:
Concentration of Na2CrO4 solution = 2.50 M
Volume of Na2CrO4 solution = 95.0 mL = 0.095 L
Number of moles of Na+ in Na2CrO4 solution = 2.50 M x 0.095 L = 0.2375 moles
Next, let's find the number of moles of Na+ in the AgNO3 solution. The equation for the reaction between Na2CrO4 and AgNO3 is:
2 Na2CrO4(aq) + 2 AgNO3(aq) → Na2Cr2O7(aq) + 2 NaNO3(aq) + 2 AgCl(s)
From the balanced equation, we can see that for every 2 moles of Na2CrO4, we get 2 moles of Na+.
Given:
Concentration of AgNO3 solution = 2.50 M
Volume of AgNO3 solution = 125 mL = 0.125 L
Number of moles of Na+ in AgNO3 solution = (2.50 M x 0.125 L) / 2 = 0.15625 moles
Now, we can add the number of moles of Na+ from both solutions to find the total number of moles of Na+:
Total moles of Na+ = 0.2375 moles + 0.15625 moles = 0.39375 moles
Finally, to find the concentration of Na+ in the mixed solution, divide the total moles by the total volume:
Concentration of Na+ = Total moles / Total volume
Concentration of Na+ = 0.39375 moles / (0.095 L + 0.125 L) = 1.08 M
Therefore, the concentration of Na+ after the two solutions are mixed together is 1.08 M.