I wanted to calculate the average equivalent weight of an unknown acid for an acid-base titration experiment. The mass of the unknown that I had to obtain is 0.2 g and the equivalents of the acid is .001046 L x N. But is this equation right? Does the equivalent mass equal to the mass of the solid acid divided by the number of equivalents of acid? If so, isn't the average eq. wt. too high?

Thank you so much for any help.

liters x normality x equivalent weight = grams.

wait, liters x normality = grams because I wanted to find the equivalent wt

Wait, im not making sense. I know liters x normality gives you the equivalents of an acid. But are you sure it's not the grams I obtained divided by the equivalents of the acid.

Of course it is but that's what I wrote in a different form.

liters x normality x eq wt = grams.
liters x normality = # equiv so
#eq x eq wt = grams and
eq wt = grams/#equivalents

I like
liters x normality x equivalent weight = grams BECAUSE it solve one equation instead of three and get the same answer.

ok, thank you very much for your time and patience. I understand, im sorry if I bothered you too much. But thanks a lot.

To calculate the average equivalent weight of an unknown acid, you are correct in using the formula:

Equivalent weight = mass of unknown acid / number of equivalents of acid

In your case, you have a mass of 0.2 g for the unknown acid and an equivalent value of 0.001046 L x N for the acid.

To calculate the number of equivalents of acid, you need to determine the concentration (N) of the acid. Once you have the concentration, you can multiply it by the volume in liters (0.001046 L) to get the number of moles, and then multiply that by the number of acidic hydrogens to get the number of equivalents. This can be done using the equation:

Number of equivalents of acid = concentration (N) x volume (L) x number of acidic hydrogens

Now, to determine the average equivalent weight, divide the mass of the unknown acid by the number of equivalents of acid. This will give you the average weight per equivalent.

Regarding your concern about the average equivalent weight being too high, it is important to consider other factors that might affect the result. Simple errors in measurements or calculations could lead to discrepancies in the final value. Make sure to double-check your values and calculations to ensure accuracy and validity of your results.

I hope this explanation helps! Let me know if you have any further questions.