Chemistry

posted by .

A few questions I don't really get and need to see the work for

A 50.0 mL sample of 0.55 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.51 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 multiplied by 10-5.

What is the pH of a solution resulting from the addition of 18.0 mL of 0.15 M HCl to 50.0 mL of 0.12 M NH3? Assume the volumes are additive. Kb of NH3 = 1.8 multiplied by 10-5.

What ratio [CH3NH3+]/[CH3NH2] is needed to prepare a buffer solution with a pH of 8.20 from methylamine, CH3NH2, and methylammonium chloride, CH3NH3Cl? Kb of CH3NH2 = 3.7 multiplied by 10-4.

What is the pH of a solution resulting from the addition of 18.0 mL of 0.15 M HCl to 50.0 mL of 0.12 M NH3? Assume the volumes are additive. Kb of NH3 = 1.8 multiplied by 10-5.

What is the molarity of a methylamine solution, CH3NH2(aq), that has the same pH as 0.0773 M NH3(aq)?

  • Chemistry -

    A few comments.
    1. Five problems in one post-- four too many.
    2. No work at all does not engender help and "I don't get it and need to see the work" is not a high incentive. Instead, post one problem and tell us what you don't understand. Surely you understand some things. Telling us what you know and what you don't understand helps us know where to start with the problem. We can help you better if we know how to explain it. Thanks.
    3. Some of the rationale for the above comes from the fact that it can take up to 30 minutes each to type in a problem with all of the exponents etc and that can easily be wasted time if you already know that part of our explanation. Again, thanks.

  • Chemistry -

    The secret to doing these titration problems is in knowing what you have at certain points in the titration. For benzoic acid titrated to the equivalence point, the acid has been converted to the salt and one has sodium benzoate. Salts, as you know, hydrolyze. If we call benzoic acid HBz and the benzoate anion Bz, then
    Bz^- + HOH ==> HB + OH^-
    Kb for Bz^- = (Kw/Ka) = (HBz)(OH^-)/(Bz^-).
    Substitute Kw, Ka, (Bz = moles/L) and (HBz)=(OH^-)=x. Solve for x, convert to pOH, then pH.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    A 50.0 mL sample of 0.51 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.40 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 10-5. what am i supposed to do with this?
  2. Chemistry

    A 50.0 mL sample of 0.42 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.40 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 10-5. I know I have to set up an ICE chart, but I keep getting …
  3. 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 …
  4. Chemistry

    A 50.0 mL sample of 0.12 M formic acid, HCOOH, a weak monoprotic acid, is titrated with 0.12 M NaOH. Calculate the pH at the following points in the titration. Ka of HCOOH = 1.8 multiplied by 10-4. What is the pH when 50.0 mL NaOH …
  5. Chemistry

    Benzoic acid (C6H5COOH)is a monoprotic weak acid with Ka=6.30*10^-5. What is the pH of a solution of benzoic acid that is 0.559M and has 2.25*10^-2M NaOH added?
  6. chemistry

    A 1.213g sample of an unknown weak monoprotic acid was titrated to the endpoint with 27.25mL of a 0.3016M NaOH Calculate: -the number of moles of base reacted. -the number of moles of acid originally present. -the molar mass of the …
  7. chemistry

    enzoic acid is a weak monoprotic acid with Ka = 6.5×10-5 M. NaOH(s) was gradually added to 1.00 L of 8.66×10-2 M benzoic acid. 1. Calculate the pH of the solution before the addition of the base. 2. Calculate the pH of the solution …
  8. chemistry

    i need help solving this Benzoic acid is a weak, monoprotic acid (Ka = 6.3 × 10−5). Calculate the pH and the percent dissociation of each of the following solutions of benzoic acid. Then use Le Châtelier’s principle to explain …
  9. chemistry

    A 15.00 mL sample of an unknown monoprotic weak acid solution is titrated with 0.35 M NaOH. The initial buret reading is 0.23 mL. The phenolphthalein indicator turns the solution light pink when the buret reads 29.58 mL. A. what volume …
  10. chemistry help

    A 15.00 mL sample of an unknown monoprotic weak acid solution is titrated with 0.35 M NaOH. The initial buret reading is 0.23 mL. The phenolphthalein indicator turns the solution light pink when the buret reads 29.58 mL. B. How many …

More Similar Questions