Hi, how do I calculate how many grams of unknown diprotic acid are needed to make 150mL of 0.750N (normal) solution?
and to make 125mL of 0.500M (molar) solution?
we have 0.5000M of NaOH..
Please help me...
a).
How many equivalents do you need? THat is L x N = 0.150 x 0.750 = ?
grams = equivalent weight x equivalents.
You don't list the diprotic acid; however, the equivalent weight of the diprotic acid = molar mass/2.
To calculate the grams of unknown diprotic acid required to make a solution, we need to follow these steps:
Step 1: Determine the molar mass of the diprotic acid
The molar mass of the unknown diprotic acid is needed to convert between moles and grams. You can find the molar mass of the compound by adding up the atomic masses of all the elements present in the chemical formula.
Step 2: Calculate the number of moles of the diprotic acid needed
To calculate the number of moles, we can use the formula:
moles = normality x volume (in liters)
For the first question:
moles = 0.750 N x 0.150 L = 0.1125 moles
For the second question:
moles = 0.500 M x 0.125 L = 0.0625 moles
Step 3: Convert moles to grams using the molar mass
Using the number of moles calculated in Step 2, we can now convert moles to grams using the molar mass of the diprotic acid.
grams = moles x molar mass
For example, if the molar mass of the diprotic acid is 100 g/mol:
For the first question:
grams = 0.1125 moles x 100 g/mol = 11.25 grams
For the second question:
grams = 0.0625 moles x 100 g/mol = 6.25 grams
Make sure to use the correct molar mass of your specific diprotic acid in the calculations.
As for the additional information about the concentration of NaOH, it doesn't appear to be directly related to the calculation of the diprotic acid. If you have any specific questions regarding the NaOH concentration or need further assistance, please let me know.