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?

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

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