Question 1: Calculate the pH at the equivalence point in titrating 0.100 M solut

Question

Question 1:

Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 8.0×102M NaOH.

For Lactic Acid (HC3H5O3), PH=?

For Sodium Hydrogen Chromate (NaHCrO4), PH=?

Question 2:

Calculate the pH at the equivalence point for titrating 0.190 M solutions of each of the following bases with 0.190 M HBr.

For Sodium Hydroxide (NaOH), ph= ?

For Hydroxylamine (NH2OH), ph=?

Aniline (C6H5NH2), Ph= ?

Question 3:

A 20.0 mL sample of 0.200 MHBr solution is titrated with 0.200 MNaOH solution. Calculate the pH of the solution after the following volumes of base have been added.

For 20.2 mL, Ph=?

Question 4:

How many moles of sodium hypobromite (NaBrO) should be added to 1.00 L of 5.5×102 M hypobromous acid (HBrO) to form a buffer solution of pH = 9.13? Assume that no volume change occurs when the NaBrO is added.

mol= ?

Explanation / Answer

This has multiple question and each question has multiple parts. I am suppose to answer only 1 question at a time. So I will answer question 1.
Question 1:
This is a reaction between weak acid and strong base. Which will form buffer.
Let weak acid be denoted by HA which stands for both lactic acid and Sodium Hydrogen Chromate

                                HA +              NaOH   --->          A- +          H2O
initial:                            0.1      (8*10^-2)
at equilibrium :    0.1- (8*10^-2)            0               8*10^-2    8*10^-2

[HA] = 0.1- (8*10^-2)   = 0.02 M

[A-] = 8*10^-2 M = 0.08 M

For lactic acid:

pKa = 3.86

use:

pH = pKa +log {[A-]/[HA]}

     = 3.86 + log {0.08/0.02}

    = 3.86 + 0.6

     = 4.46

For For Sodium Hydrogen Chromate:

pKa = 6.5

use:

pH = pKa +log {[A-]/[HA]}

     = 6.5 + log {0.08/0.02}

    = 6.5 + 0.6

     = 7.1

Sorry I can't answer any more question. Please ask them as different questions

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