Concentration of HCl used 0.500 M Vol HCl added (mL) 49.67 mL Concentration of NaOH 0.453 M Total moles of HCl added Moles HCl = Molarity x volume in liters Initial buret reading (NaOH) 25.65 mL Final buret reading (NaOH) 49.72 mL Vol NaOH added (mL) Final -Initial buret readings (NaOH) Moles of NaOH used Moles NaOH used = Molarity x volume in L Moles of HCl neutralized by NaOH Moles of HCl neutralized by NaOH = moles NaOH (1mole acid/1 mole base) Moles of HCl neutralized by antacid Moles of HCl neutralized by the antacid = total moles HCl added – moles HCl neutralized by NaOH Moles of HCl consumed per gram of antacid Moles of HCl consumed per gram of antacid= moles HCl neutralized by antacid/ grams of antacid
|
|
Trial 1 |
Calculations |
|
Weight of Antacid
|
1.541 g |
|
|
Concentration of HCl used |
0.500 M |
|
|
Vol HCl added (mL) |
49.67 mL |
|
|
Concentration of NaOH |
0.453 M |
|
|
Total moles of HCl added |
|
Moles HCl = Molarity x volume in liters |
|
Initial buret reading (NaOH) |
25.65 mL |
|
|
Final buret reading (NaOH) |
49.72 mL |
|
|
Vol NaOH added (mL) |
|
Final -Initial buret readings (NaOH) |
|
Moles of NaOH used |
|
Moles NaOH used = Molarity x volume in L |
|
Moles of HCl neutralized by NaOH |
|
Moles of HCl neutralized by NaOH = moles NaOH (1mole acid/1 mole base) |
|
Moles of HCl neutralized by antacid |
|
Moles of HCl neutralized by the antacid = total moles HCl added – moles HCl neutralized by NaOH |
|
Moles of HCl consumed per gram of antacid |
|
Moles of HCl consumed per gram of antacid= moles HCl neutralized by antacid/ grams of antacid |
calculate the moles of HCl nuetralized by the NaOH from the titration data
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
