EXPERIMENT 5 PRE-LABORATORY ASSIGNMENT 1. a) (2 points) Write the net ionic equa

Question

EXPERIMENT 5 PRE-LABORATORY ASSIGNMENT 1. a) (2 points) Write the net ionic equation for hydrochloric acid reacting with solid sodium carbonate b) (3 points) On average, one tablet of a certain antacid contains 500.0 mg sodium carbonate. If one tablet has a cost of 15 cents, calculate the cost of neutralizing hydrochloric acid in dollars per mole of HCI. (S/mol) Answer: 2. a) (3 points) If 0.900 g sodium carbonate is added to 325. mL of 0.150 M HCI, calculate the molarity of H' in the solution after the reaction above goes to completion. Answer: b) (2 points) What volume of 0.950 M KOH would be required to neutralize the remaining HCI solution in part (a)? (do not use M1V1M2V2) Answer:

Explanation / Answer

a)-All carbonates and bicarbonates are instantly decomposed by acids with brisk effervescence of CO2, along with the formation of the salt of the acid.

                   Na2CO3(s) + 2HCl(aq) = 2NaCl(aq) + CO2(g) + H2O(l)

b)-here chemical Reaction: Na2CO3(s)+2HCl(aq)--->2 NaCl(aq)+H2O(l)+CO2(g)

500.0 mg Na2CO3 x 1 g/1000 mg x 1 mol Na2CO3/105.984 g Na2CO3 x 2 mol HCl/1 mol Na2CO3 x 2 mol H+/2 mol HCl = 0.009435386

15 cents x 1dollar/100 cents= $0.15

$0.15/0.009435386= $16/1 mol HCl

c)-2 moles of HCl react with every mole of Na2CO3. So you need to calculate how many moles of H are in 325ml of 0.15M HCl. Double the number of moles of Na2CO3 you have to get the number of moles of H it will react with. Take this number of moles away from the total number of moles of HCl you started with. Whatever is left is the number of moles of H+ you have left over. To get the molarity you put this number of moles into 325ml and calculate the molarity.

d)-1.87 g NaHCO3 = 1.87g * (mole/84g)=
0.0222 mole

350ml*0.1moleH+/1000ml= 0.035 mole H+

H+ left = 0.035 - 0.0222 = 0.0128 moles

the molarity is in 350ml soln = 0.0128/.35 = 0.036M

B) xml *0.950moleOH/1000ml = 0.0128 mole H+

solving x = 13.47 ml

you'll need 13.47 ml of 0.95M NaOH soln to neutralize whats left.

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