# Titration Curve Question- General Chemistry?

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This question goes into an online homework program, and I cannot seem to get it right.

A 15.0 mL sample of 0.100 M Ba(OH)2 is titrated with 0.125 M HCl. Calculate the pH for at least 5 points and create a titration curve.

I have:

At 0 mL HCl::

Solution is all OH-, so pOH= -log(.1)= 1. pH = 13

But then when I try 5 mL HCl, I get:

.005 L x .125 mol/L = 6.25×10^-4 mol HCl.

.003 mol – (6.25 x 10^-4)= .002375 mol OH-

.002375 mol/ .02 L= .11875 M OH-

pOH= .925

pH= 13.07

What am I doing wrong?

• First write a balanced equation:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

1mol Ba(OH)2 reacts with 2 mol HCl to produce 1 mol BaCl2

The BaCl2 is the salt of a strong acid / strong base reaction, so at equivalence pH of the titration will be 7.00

Calculate the volume of 0.125M HCl required for equivalence:

mol Ba(OH)2 in 15.0mL of 0.100M solution = 15/1000*0.100 = 0.0015 mol Ba(OH)2

This will require 0.0030mol HCl

Volume of 0.125M HCl that contains 0.0030mol = 0.0030/0.125*1000 = 24.0mL

In order to draw the graph ( and hopefully teach ourselves something important) let us calculate the pH after the following additions of HCl:

1) 0mL

2) 12.0mL

3) 23.9mL

We already know 24.0mL , pH = 7.00

4) 24.1mL

5) 36.0mL

Now calculate the pH of the Ba(OH)2 solution before any addition of HCl

Ba(OH)2 dissociates : Ba(OH)2 ↔ Ba2+ + 2OH-

Therefore in the Ba(OH)2 solution [OH-] = 0.20M

pOH = -log 0.200

pOH = 0.699

pH = 13.30

Calculate the pH at addition of 12.0mL HCl:

Remember that you initially had 0.0015 mol of Ba(OH)2 in the 15.0mL sample .This is 0.0030 mol of OH- ions. By adding 12.0mL of the HCl, you neutralise half of these OH- ions. You have 0.0015 mol OH- ions in 15+12 = 27mL of solution

Molarity of the OH- ions = 0.0015/0.027 = 0.05556M

[OH-] = 0.05556M

pOH = -log 0.05556

pOH = 1.26

pH = 14.00-1.26 = 12.74

At 12.0mL HCl addition, pH = 12.74

Now calculate the pH of the solution when 23.9mL of HCl have been added:

You originally had 0.0015 mol Ba(OH)2 or 0.0030 mol of OH- ions in solution

By adding 23.9mL of the acid you neutralise: 23.9/24.0*0.0030 = 0.0029875 mol of OH- ions

You have 0.0030 – 0.0029875 = 0.0000125 mol OH- ions remaining, dissolved in 15+23.9 = 38.9 mL solution

[OH-] = 0.0000125/0.0389 = 0.000321M

pOH = – log 0.000321 = 3.5

pH = 14.0-3.5 = 10.5

This I think is the most important and fascinating aspect of this – You have added 99.5% of the acid and the pH of the solution is 10.5

We know that if you add a further 0.1mL of the HCl, you will have complete neutralisation, and pH will be 7.00.

I am not going to do all the calculations, but you can do this yourself and you will find that when you add 24.1mL of the acid, the pH will be 3.5 . When you have added a further 12.0mL of the acid, you will have 0.0015 mol HCl dissolved in 51.0mL of solution:

[H+] = 0.0015/0.051 = 0.029 M

pH = -log 0.029

pH = 1.53

0ml HCl : pH = 13.3

12.0mL HCl : pH = 12.74

23.9mL HCl : pH = 10.5

24.0mL HCl : pH = 7.00

24.1mL HCl : pH = 3.5

36mL HCl : pH = 1.53

Now draw your graph: The most important thing to note is – because you are reacting a strong base / strong acid, is the very large change in pH at the equivalence point. The graph is vertical. I am sure that this is what your teacher wants you to show ( and understand).

• since its strong acid+strong base the product is H2O and salt, you know its going to be a 1:1 reaction, and you have the concentrations.

Ba(OH)(aq) + HCl(aq) –> H2O(l) + BaCl(aq) *always include the equation, (aq) means aqueous, pretty much anything in solution

the first point is good, the last point should be pH=7. add 3mL each point not 5, the equivalence point should be reached at 12mL HCl. 1/1.25=0.8 and 0.8*15mL=12mL, so 0ml, 3ml,6ml,9ml, and 12ml.

sorry in advance for the crazy way to write it, easier on paper obviously

2nd point should go;

1.510^(-3)mol OH – 3.7510^(-4)mol H+ = 1.12510^(3-) OH /makes a little salt and water

pOH= -log(1.125*10^(3-) OH / .018L = 1.2

pH = 14 – 1.2 = 12.8

same steps for the next points;

point 3 6mL HCl pOH=1.447 pH=12.5

point 4 9mL HCl pOH=1.8 pH=12.2

point 5 12mL HCl pH=pOH=7 * you might want to add a point like 11mL or graph it with .01mL increments, the reason behind this is that when titrating strong acids with strong bases the graph goes really steep through the equivalent point. you’ll see this when you use a burret, like one drop and the indicator used will change, in our labs we usually phenol red i believe and its like a faint pink color your aiming for, either way good luck.

Source(s): Im in 2nd year science lab tech, lots of lab experience titrating

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