i measured out 2 different tared mass of TUMS sample. One is 0.1554g and the other 0.1546g.
Then i did two titrations with the conclusion of one being 17.91mL of NaOH and the other 17.47mL
my question is: "if you consumed two 1.00g tablets of TUMS, how many moles of acid would be neutralized"??
this is a back titration of standardized HCl with Sodium Hydroxide solution
You didn't provide the volume or M of the HCl and you must have that to answer this question.
You added an excess of HCl. How much HCl did you add? moles HCl = M x L = ?
Then you back titrated with NaOH. mols NaOH = M x L = ?
The difference in moles = moles of acid consumed by the TUMS. I would calculate each titration separately and divide the final result of each by the mass of the original sample which should give you moles H^+/grams and that times 2 will give moles/2 grams.
the volume and M of HCl is 0.1312M and 150mL
how do i find then excess of HCl
What volume of 0.1125 M K2Cr2O7 would be required to oxidize 48.16 mL of 0.1006 M Na2SO3 in acidic solution? The products include Cr3+ and SO42- ions.
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To determine the number of moles of acid neutralized by two 1.00g tablets of TUMS, we need to consider the back titration of standardized HCl with sodium hydroxide (NaOH) solution.
First, let's calculate the moles of NaOH used in each titration:
For the first titration:
Volume of NaOH solution used = 17.91 mL = 0.01791 L
Molarity of NaOH solution (assuming it's standardized) = given or known value (let's say it is 0.1 M)
Moles of NaOH used = Volume (L) x Molarity (mol/L)
= 0.01791 L x 0.1 mol/L
= 0.001791 mol
For the second titration:
Volume of NaOH solution used = 17.47 mL = 0.01747 L
Moles of NaOH used = Volume (L) x Molarity (mol/L)
= 0.01747 L x 0.1 mol/L
= 0.001747 mol
Now, we need to determine the number of moles of acid neutralized in each titration. Since it's a back titration, we can assume that the moles of acid neutralized equal the moles of NaOH used.
Therefore, the number of moles of acid neutralized in the first titration = 0.001791 mol
And the number of moles of acid neutralized in the second titration = 0.001747 mol
To calculate the total number of moles of acid neutralized when two 1.00g tablets of TUMS are consumed, we need to find the molar mass of the acid present in TUMS. It is commonly calcium carbonate (CaCO3), which contains one mole of acid.
The molar mass of CaCO3 = (40.08 g/mol of Ca) + (12.01 g/mol of C) + (3 x 16.00 g/mol of O)
= 100.09 g/mol
Now we can calculate the number of moles of acid neutralized by two tablets of TUMS:
Mass of one tablet of TUMS = 1.00 g
Molar mass of TUMS (CaCO3) = 100.09 g/mol
Moles of acid neutralized by one tablet = Mass (g) / Molar mass (g/mol)
= 1.00 g / 100.09 g/mol
= 0.009995 mol
Since we have two tablets, the total number of moles of acid neutralized when both tablets are consumed = Moles per tablet x Number of tablets
= 0.009995 mol x 2
= 0.01999 mol
Therefore, consuming two 1.00g tablets of TUMS would neutralize approximately 0.01999 moles of acid.