How do I approach this Chemistry problem?

[Johnny_L.A]

Please don’t post the answer to the problem. I’m just looking for guidance on how to approach it.

A 40.0 mL volume of 0.100 M NaOH is titrated with 0.0500 M HCl. Calculate the pH after addition of the following volumes of acid.

Part A: 64.5

So I’m adding 0.0645 L of 0.0500 M HCl. 0.0645 * 0.05 = 0.003225 moles of HCl. The pH of 0.003225 mol HCl = -log(0.003225) = 2.49. pH + pOH = 14, so pOH = 11.51. So I enter 11.51 and get this: ‘Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.’

I also tried it this way:

40.0 mL * 0.100 mol/L = .004 mol NaOH
Add 64.5 mL 0.0500 M HCl = 0.003225 mol HCl

Initial moles NaOH = 0.004000
Mol HCl added = 0.003225
Mol NaOH remaining = 0.000775

Molarity = 0.000775 / (40 ml + 64.5 ml) = 7.416267943x10[sup]-6[/sup]
ph = -log(7.416x10[sup]-6[/sup]) = 5.13

11.51 is ‘close, but no cigar’, and 5.13 isn’t even that close.

I’ll find out where I’m going wrong on Tuesday, so I’m not desperate for an answer. But I’ve got a horrible cold, and I’m frustrated; so if I can get a hint I can put what’s left of my mind at east for a few days.

[Raygun99]

@Johnny_L.A

40.0 mL * 0.100 mol/L = .004 mol NaOH
Add 64.5 mL 0.0500 M HCl = 0.003225 mol HCl

Initial moles NaOH = 0.004000
Mol HCl added = 0.003225
Mol NaOH remaining = 0.000775

Molarity = 0.000775 / (40 ml + 64.5 ml) = 7.416267943x10[sup]-6[/sup]
ph = -log(7.416x10[sup]-6[/sup]) = 5.13

You’ve calculated the pOH here. If you think about it, you have more moles of [OH-] than [H+], so your pH has to be above 7.

[Bill_Door]

You’ve got the moles right, there are 0.000775 in 0.1045 liters, making the molarity 7.4163 X10^-3, not 10^-6. I think, it’s been a while. Your pOH ends up 2.13, and your pH 11.87.

[Johnny_L.A]

@Bill_Door

You’ve got the moles right, there are 0.000775 in 0.1045 liters, making the molarity 7.4163 X10^-3, not 10^-6.

:smack:

Thank you!

@Bill_Door

Your pOH ends up 2.13, and your pH 11.87.

I didn’t read that far. It was only after I recalculated and started this reply that I saw it. :smack:

[johnpost]

it’s pretty basic.

your volume is too small. and what is there is OH.

[johnpost]

colds and frustration destroy calculations.

[Johnny_L.A]

@johnpost

it’s pretty basic.

your volume is too small. and what is there is OH.

Things that get me every time:
[ul][li]Signs[/li][li]Forgetting (or mis-signing) exponents in scientific notation[/li][li]Units[/li][li]Getting a wrong number lodged in my head. (Example: 1.4x10[sup]-14[/sup] instead of 1.0x10[sup]-14[/sup].)[/ul][/li]
EDIT:

@johnpost

colds and frustration destroy calculations.

I friggin’ hate being sick!

.

[johnpost]

@Johnny_L.A

Things that get me every time:
[ul][li]Signs[/li][li]Forgetting (or mis-signing) exponents in scientific notation[/li][li]Units[/li][li]Getting a wrong number lodged in my head. (Example: 1.4x10[sup]-14[/sup] instead of 1.0x10[sup]-14[/sup].)[/ul][/li]

it also might help to convert all values to a single order of magnitude (liter, mole) before doing calculations. keep everything in scientific notation.

this can reduce some of the clutter that can obscure errors.

later with practice you might be able do do calculations with values over 9 orders of magnitude in your head but you need to have concepts and formulas solid in your head first.

[Johnny_L.A]

Still sick, meds are making me hazy. Just thinking ‘out loud’…

Consider the titration of 100.0 mL of 0.016 M HOCl (Ka=3.5×10[sup]-8[/sup]) with 0.0400 M NaOH. Calculate the pH after the addition of 10.0 mL of 0.0400 M NaOH.

I know the answer is pH = 6.98.

The reaction is HOCl + NaOH <==> NaOCL + H2O.
I start with 100 mL * 1 L/1000 mL * 0.016 mol HOCl = 0.0016 mol HOCl.
I add 10 mL * 1 L/1000 mL * 0.0400 mol NaOH = 0.0004 mol NaOH
I should have 0.0016 - 0.0004 = 0.0012 mol HOCl remaining.

pH = pKa + log([NaOH]/[HOCl] = -log(3.5x10[sup]-8[/sup]) + log(0.0004/0.0012) = 6.98.

Yay! Got it right!

Now I have to calculate the pH halfway to the equivalence point, and the pH at the equivalence point. But those can wait until after breakfast.

[Johnny_L.A]

@Johnny_L.A

Now I have to calculate the pH halfway to the equivalence point

Halfway to the equivalence point, I’ve added enough NaOH to use up/leave half of the HOCl. So the concentration of NaOH and the concentration of HOCl are both the same. Since pH = pKa + log([base]/[acid]), and since [base]/[acid] = 1, and since log(1) = 0, then pH = pKa. So pH = -log(3.5x10[sup]-8[/sup] = 7.46.

@Johnny_L.A

and the pH at the equivalence point.

Since we start with 100 mL 0.016 M = 0.0016 moles HOCl, we need 0.0016 moles of NaOH at the equivalence point. 0.0016 mol NaOH * 1 L/0.04 mol = 0.04 L = 40 mL. The equivalence point is where 40 mL NaOH has been added to the solution.

OK, now I’m stuck. I have 140 mL = 0.14 L of solution. Since all of the HOCl has been reacted, I have 0 moles of that. I have 0.0016 mol Na[sup]+[/sup] floating around. I know that at the equivalence point the solution is basic.

How do I use the numbers I have to find the pH? I know it’s staring me in the face, but I’m either using the wrong equation or else using the numbers wrong in the right equation.

[Bill_Door]

@Johnny_L.A

Halfway to the equivalence point, I’ve added enough NaOH to use up/leave half of the HOCl. So the concentration of NaOH and the concentration of HOCl are both the same. Since pH = pKa + log([base]/[acid]), and since [base]/[acid] = 1, and since log(1) = 0, then pH = pKa. So pH = -log(3.5x10[sup]-8[/sup] = 7.46.
(snip)

Wait a minute, now I’m confused. Halfway to the equivalence point you don’t have equal concentrations of NaOH and HOCl, you have no NaOH unreacted, and half of your HOCl has reacted, leaving 0.0008 moles of HOCl. Then you use the pKa to determine the pH of 120 ml of a solution containing 0.0008 moles HOCl, because it takes 20 ml of 0.0400M NaOH to supply 0.0008 moles. Maybe I’m overtired.

[Johnny_L.A]

@Bill_Door

Wait a minute, now I’m confused.

You’re confused! I’m confused and sick!

From a different weak acid-strong base reaction:

After addition of 20.0 mL of 0.100 M NaOH, we have added 2.00 mmol of NaOH, which is enough OH[sup]-[/sup] to neutralize exactly half of the 4.00 mmol of CH[sub]3[/sub]CO[sub]2[/sub]H present initially. Neutralization gives a buffer solution that contains 2.00 mmol of CH[sub]3[/sub]CO[sub]2[/sub][sup]-[/sup] and 4.00 - 2.00 mmol of CH[sub]3[/sub]CO[sub]2[/sub]H. Consequently, the [base]/[acid] ratio is 1.00 and pH = pKa:
pH = pKa _ log([base]/[acid] = pKa = 4.74
Halfway to the equivalence point is a special point in the titration of any weak acid with a strong base because at that point the pH always equals the pKa of the weak acid.

[asterion]

Sometimes it can help to use units that reduce the amount of decimal points. For instance, 0.4 mmol (that’s milli-mole) instead of 0.0004 mol. That’s assuming you don’t want to use scientific notation, of course.

Anyway, as a working chemist my approach to “what’s the pH of this titration?” is to stick the pH probe in it and write down the number.