Post a New Question


posted by .

A titration involves titrating 0.4 M NaOH into a solution of 0.2 M H3PO4.
a.) Calculate the volume of NaOH that will be required to reach the first equivalence point.
b.) Calculate the volume of NaOH required to reach the second equivalence point.
Other information: 10 ml H3PO4 was added to 250 ml of deionized water, and solution was then titrated with NaOH.

  • Chemistry -

    Is that 10 mL of the 0.2M H3PO4 added to 250 mL H2O to make 260 mL solution? If you titrate the entire solution, which seems unlikely, then you will have
    H3PO4 ==> NaH2PO4 + H2O

    mols H3PO4 = 0.010L x 0.2M = 0.002
    mols NaOH required = 0.002 since 1 mol H3PO4 = 1 mol NaOH.
    Then M NaOH = mols NaOH/L NaOH or
    L NaOH = mols NaOH/M NaOH = 0.002/0.4 = approx 0.005L or 5 mL

    For part B you are titrating the second H do it will take twice that or abouat 10 mL.

    If I've interpreter the problem incorrectly please rephrase the question so it is clear what is being titrated.

  • Chemistry -

    Oh, sorry. It was supposed to be 50 ml of water instead of 250 ml.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    acid: HCI 0.10 M Base:NaOH O.050 M 3 ml of HCI was used. Calculate the volume of base required to reach the equivalence point of this titration. Somewhere in the finding the solution, i have to use the fact that Moles HCl = moles NaOH. …
  2. college chemistry

    In class, we discussed the titration of a 50.0 mL sample of 0.100 M HCL with a 0.100 M solution of NaOH. How would this system change if we used a 0.100 M solution of Sr(OH)2 instead of NaOH?
  3. Chemistry

    In this case, the inflection point, and equivalence, occurs after 23.25mL of 0.40 M NaOH has been delivered. Moles of base at the equivalence point can be determined from the volume of base delivered to reach the equivalence point …
  4. College Chemistry

    1) A 25mL sample of the .265M HCI solution from the previous question is titrated with a solution of NaOH. 28.25mL of the NaOH solution is required to titrate the HCl. Calculate the molarity of the NaOH solution. 2) A 1.12g sample …
  5. chemistry

    What volume of 0.115M NaOH is required to reach the equivalence point in the titration of 25.00 mL of 0.200M of HF?
  6. chemistry

    a sample of 20.0 mL of 0.100 M HCN (Ka=6.2*10^-10) is titrated with 0.150 M NaOH. a) what volume of NaOH is used in this titration to reach the equivalence point?
  7. Chemistry

    A solution contains 25 mmol of H3PO4 and 10. mmol of NaH2PO4. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH?
  8. chemistry

    11.During an acid-base titration, 25 mL of NaOH 0.2 M were required to neutralize 20 mL of HCl. Calculate the pH of the solution for each of the following: 12.Before the titration. 13.After adding 24.9 mL of NaOH. 14.At the equivalence …
  9. Physical Chemistry

    If 20.0mL of 0.122M NaOH are required to reach the first equivalent point of a solution of citric acid(tripotic acid H3C6H5O7).How many mL of NaOH in total are required to reach the second equivalence point?
  10. Chemistry

    You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate with a 0.1105 M NaOH. If you add 20.00 mL of HCCOOH to the beaker before titrating, approximately what volume of NaOH …

More Similar Questions

Post a New Question