Question

A student was titrating 25.00 mL of a monoprotic basic solution with a monoprotic acidic solution...

A student was titrating 25.00 mL of a monoprotic basic solution with a monoprotic acidic solution that was 0.281 M. The student ran out of the acidic solution after having added 32.46 mL, so she borrowed a solution that was labeled as 0.317 M. An additional 11.5 mL of the second solution was needed to complete the titration. What was the concentration of the basic solution?

I know how to solve simple titration but how do I solve a multistep one like this?

Homework Answers

Answer #1

you should mention the borrowed solution is mono protic acid or not

ok i am assuming the borrowed solution is mono protic acidic solution

firs convert every thing in to the moles

intiall he added 32.46 mL of 0.281 M acidic solution

no of moles of acid = Molarity x volume in liters = 0.281 M x 0.03246 L = 0.009131 mol

no of moles of acid you borrowed is = 0.317M x 0.0115 L = 0.0036455 mol

total moles of acid = 0.009131 + 0.0036455 = 0.01277 mol

since he mentioned both are mono protic acid and mono protic base the reaction will be

HA + BOH ------> AB + H2O

menas one mole of acid will react with one mole of base

accordingly 0.01277 mol will react with 0.01277 mole of base

now you have a no of moles of Base and volume of the base find the concentration using the formula

Molarity = moles / volume in liters = 0.01277 mol / 0.025 L

= 0.511 M

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A solution of MgCl2 is standardized by titrating 25.00 mL aliquots of the solution with a...
A solution of MgCl2 is standardized by titrating 25.00 mL aliquots of the solution with a 0.01162 M EDTA solution, using eriochrome black T as the indicator. The volumes of titrant required for the titration of three aliquots are given in the table below. Determine the concentration of the MgCl2 solution and its standard deviation. (4 pts) aliquot Volume EDTA (mL) 1 39.83 2 40.03 3 39.97
A student titrated a 25.00 mL sample of a solution containing an unknown weak, diprotic acid...
A student titrated a 25.00 mL sample of a solution containing an unknown weak, diprotic acid (H2A) with NaOH. If the titration required 17.73 mL of .1036 M NaOH to completely neutralize the acid, calculate the concentration (in M) of the weak acid in the sample.
Describe a titration when a basic solution is added to an acidic solution and explain whether...
Describe a titration when a basic solution is added to an acidic solution and explain whether this is proper technique. Include any additional precautions that might be necessary during this titration.
student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution...
student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.25 mL of the weak acid solution has been added to 50.0 mL of the 0.100 M solution, the pH of the resulting solution is 10.50. Calculate the original concentration of the solution of weak acid. Concentration =__ M
During a titration experiment, a student titrated 25.00 mL of a 0.100 M sodium hydroxide solution...
During a titration experiment, a student titrated 25.00 mL of a 0.100 M sodium hydroxide solution with 5.10 mL of a 0.250 M sulphuric acid solution. What is the pH of the resulting solution?
A student pipetted 25.00 mL of a stock solution that was 0.1063 M HCl into a...
A student pipetted 25.00 mL of a stock solution that was 0.1063 M HCl into a 100.00 mL volumetric flask and diluted the solution to the volumetric flask calibration mark with deionized water. Calculate the concentration of the diluted solution.
A student titrated 38.00 mL of a 0.522 M sodium hydroxide solution with 25.00 mL of...
A student titrated 38.00 mL of a 0.522 M sodium hydroxide solution with 25.00 mL of a 0.785 M hydrochloric acid solution. Which is the limiting reagent? How many mL of excess reactant will remain?
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 0.03828 M EDTA solution. The solution is then back titrated with 0.02192 M Zn2 solution at a pH of 5. A volume of 15.73 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 0.05231 M EDTA solution. The solution is then back titrated with 0.02324 M Zn2 solution at a pH of 5. A volume of 20.98 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 0.04127 M EDTA solution. The solution is then back titrated with 0.02003 M Zn2 solution at a pH of 5. A volume of 16.44 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...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT