I need help with the calculations for parts C.1, C.2, and C.3. Thank you. ab Sec

Question

I need help with the calculations for parts C.1, C.2, and C.3. Thank you.

ab SecName A. Dissolving the Antacid 1. Mass of flask (8) 2. Mass of flask + antacid sample (g)32 3. Mass (or tared mass) of antacid Trial I Trial 2 sample (g) 4. Total volume of HCI added (mL.) 5. Molar concentration of HCl (moV/L) B. Analyzing the Antacid Sample 1. Molar concentration of NaOH mo/L) 2. Buret reading. initial (ml.) 3. Buret reading. final (omL) 4. Volume of NaOH (mL) C. Calculations 1. Moles of HCl added, total (mol) 2. Moles of NaOH added (mol) 3. Moles of base in antacid sample (mol) mol base in antacid mass of antacid sample (molg) mol base in antacid (mo Average mass of antacid sample "Show calculation(s) for Trial(s) 1 on the next page.

Explanation / Answer

Moles of acid and base can be calculated by using their respective concentration and volume used.

FOR MOLES OF ACID

Volume of acid used = 25 mL

Concentration of acid = 0.130 M

Now, 0.130 M (0.130 mol/L) solution means

1000 mL solution has HCl = 0.130 mole

25 mL solution has HCl = 0.130 × 25/1000 mol = 0.00325 mol

So, moles of HCl = 0.00325

Similarly, FOR BASE NaOH

Volume used = 15.23

Concentration = 0.100 M(0.100 mol/L)

Again, 1000 mL solution has NaOH = 0.100 mol

15.23 mL has NaOH = 0.100 × 15.23/1000 =0.001523 mol

So, moles of NaOH = 0.001523 moles

FOR BASE IN ANTACID

As we know when acid and base are mixed, then reaction undergo neutralisation, and number of moles of acid is equal to number of moles of base.

Since antacids are basic in nature because these are used to neutralise the extra acid produced in our body.

Now, the moles of acid and bases calculated above are not equal because base moles from ANTACID will also contribute to neutralise the acid HCl. SO,

Moles of acid = moles of NaOH + moles of base from antacid

0.00325 = 0.001523 + moles of base in antacid

Moles = (0.00325 - 0.001523) mol = 0.001727 mol

So, moles of base in antacid = 0.001727

  

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