[Geddoe]

You have 0.8744 g of an unknown monoprotic acid, HA, which reacts with NaOH according to the
balanced equation
HA + NaOH -> NaA + H2O
If 45.56 mL of 0.1148 M NaOH is required to titrate the acid to the equivalence point, what is the molar
mass of the acid?

I'm a bit confused by this. So I know:
M = n / v

n = M * v, thus:

(0.1148 M NaOH * 45.56mL) * (1 L / 1000mL) = 0.005230288 mole NaOH

where can I take it from here?

[cliverlong]

1. Look at your units

mol / mL  * mL / L = mol

Look at what you have written.

2. Once you have the correct number of moles, how do you use the balanced equation? What could you determine about the other reactants and products if you calculated you had 0.3 moles of NaOH?

[Geddoe]

1. Look at your units

mol / mL  * mL / L = mol

Look at what you have written.

2. Once you have the correct number of moles, how do you use the balanced equation? What could you determine about the other reactants and products if you calculated you had 0.3 moles of NaOH?

Well I know the ratio, thus I can write:

0.005230288 mole NaOH * (1 mole HA / 1 mole NaOH) = 0.005230288 mole HA


But I'm still confused now.

What could you determine about the other reactants and products if you calculated you had 0.3 moles of NaOH?

I'm not sure? Where did 0.3 mole NaOH come from?

[cliverlong]

Well I know the ratio, thus I can write:

0.005230288 mole NaOH * (1 mole HA / 1 mole NaOH) = 0.005230288 mole HA

You are assuming 0.005230288 mole NaOH is correct. I am saying your error is earlier than that.

I restate 1. Look at your units

It is true that

mol / mL  * mL / L = mol

That is not what you have written. I believe what you have written is incorrect and hence your calculation is incorrect.

Look at the units.

But I'm still confused now.

What could you determine about the other reactants and products if you calculated you had 0.3 moles of NaOH?

I'm not sure? Where did 0.3 mole NaOH come from?

IF, I wrote IF - as a prompt to the next step of the calculation. I think the popular term for the method you need is stoichiometry , otherwise known as mole ratios, to do the second part of the question.

However, if I am correct, you need to get the correct number of moles of NaOH first.

[DrCMS]

You are assuming 0.005230288 mole NaOH is correct. I am saying your error is earlier than that.

I think you'll find that 0.005230288 is correct, why do you think it is wrong?

Well I know the ratio, thus I can write:

0.005230288 mole NaOH * (1 mole HA / 1 mole NaOH) = 0.005230288 mole HA


But I'm still confused now.

You are correct so far you've calculated the moles of acid and as you are given the weight you can calculate the molecular weight of the acid.

[cliverlong]

I think you'll find that 0.005230288 is correct, why do you think it is wrong?

I was completely wrong.

I was focussed on the way the following equation was written

(0.1148 M NaOH * 45.56mL) * (1 L / 1000mL) = 0.005230288 mole NaOH

The way I write such calculations is:

0.1148 [mol / dm³] NaOH * 1  / 1000 [L / mL] * 45.56 [mL] NaOH

 = 0.005230288 [mol] NaOH

to make the conversion of units more explicit to me.

However, the original calculation by the original poster was correct and I was incorrect.

Clive