Question

In: Chemistry

A 25.00 mL aliquot of solution has a concentration of 3.813 x 10-5​ M, what is...

A 25.00 mL aliquot of solution has a concentration of 3.813 x 10-5​ M, what is the concentration of the 50.0 mL solution the aliquot was taken from?

Solutions

Expert Solution

before dilution                                                                 after dilution

M1 = 3.813*10-5 M                                                         M2 =

V1   = 25ml                                                                       V2 = 50ml

                M1V1    =   M2V2

                   M2     = M1V1/V2

                             = 3.813*10-5 *25/50    = 1.9065*10-5 M >>>> answer

queen_honey_blossom answered 1 year ago
Next > < Previous

Related Solutions

What is the concentration of Cu in a solution if a 25.00 mL aliquot reacted with...
What is the concentration of Cu in a solution if a 25.00 mL aliquot reacted with an excess of KI that requires 15.64 mL of 39.94 mM Na2S2O3 solution to titrate the liberated iodine?
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03742 M EDTA solution. The solution is then back titrated with 0.02190 M Zn2 solution at a pH of 5. A volume of 20.48 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05893 M EDTA solution. The solution is then back titrated with 0.02306 M Zn2 solution at a pH of 5. A volume of 19.89 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04503 M EDTA solution. The solution is then back titrated with 0.02327 M Zn2 solution at a pH of 5. A volume of 22.80 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05832 M EDTA solution. The solution is then back titrated with 0.02380 M Zn2 solution at a pH of 5. A volume of 18.49 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
25.00 mL of 0.185 M HCN (Ka = 4.90 x 10^-10) was titrated with 18.50 mL...
25.00 mL of 0.185 M HCN (Ka = 4.90 x 10^-10) was titrated with 18.50 mL of Ca(OH)2. What is the pH of the original HCN solutuon? What is the pH after 9.25 mL of base have been added?
A 25.00 mL solution of 0.100 M CH3NH2 is titrated with 0.1331 M HCl. What is...
A 25.00 mL solution of 0.100 M CH3NH2 is titrated with 0.1331 M HCl. What is the pH at the equivalence point of the titration? Species (K values) CH3NH2 (Kb = 3.74x10-4) CH3NH3+ (Ka = 2.70x10-11)
a)What is the pH of a solution prepared by adding 25.00 ml of 0.0993 M sodium...
a)What is the pH of a solution prepared by adding 25.00 ml of 0.0993 M sodium hydroxide to 30.00 mL of 0.0553 M acetic acid? There will be a reaction. What two major species remain and how much of each? b) What is the pH of 9.93x10^-8 M potassium hydroxide? c) What is the volume of 0.579 M sodium hydroxide required to reach the equivalence point of a titration of 10.0 mL of 5.00% (by mass) acetic acid? d) What...
A 0.1000 M NaOH solution was employed to titrate a 25.00-mL solution that contains 0.1000 M...
A 0.1000 M NaOH solution was employed to titrate a 25.00-mL solution that contains 0.1000 M HCl and 0.0500 M HOAc. a) Please determine the pH of the solution after 27.00 mL of NaOH is added. Ka, HOAc = 1.75*10-5 b) what's the solution pH at the second equivalence point?
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate 1) the concentration of the acid in the original solution, 2) the pH of the original HCl solution and the original NaOH solution, 3) the pH after 10.00 mL of NaOH have been added, 4) the pH at the equivalence point, and 5) the pH after 25.00 mL of NaOH have been...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT