Question

how do I calculate the isolated moles and molarity of the 0.2M KIO3?

0.1 M Sulfuric Distilled Solution A Mixturel Mixture II Mixture III Mixture IV 0.20 M KIO3 Solution 5 mL 5 mL 10 mL 10 mL Acid 0 mL 0.5 mL 0 mL 0.5 mL water 15 ml 14.5mL 10 mL 9.5 mL Lalculations: I. (6pt) Calculate the total number of moles of iodate ion present in each 0.2 M KlO3 solution. Calculate the molarity of iodate ion in 40 mL of solution for each mixture. (1000 mL = 1L) Solution A Volume of 0.2 M Moles of iodate ion Molarity of iodate ion in 40 ml solution KIO3 Mixture I 5 mL Mixture II 5 mL Mixture III 10 mL Mixture IV 10 mL I1. (8pt) Calculate Rate for each mixture. RATE EXPERIMENT: (16pt) - 3. Dete rmining reaction rate for the iodine clock experiment. (You can collect your data with o lab partner. Perform the calculations by yourself.) a given reaction, potassium iodate KIO, and sodium metabisulfite Na S20s react to form iodine. The starch solution serves as an indicator of the end of the reaction by forming a deep-blue colored starch-iodine complex. In the presence of hydrogen ions (H") from the added sulfuric acid, the iodide ions react with excess date ions to produce iodine (12). This is the kinetically slow step in the overall reaction: 6H'(aq) + 51 (aq) + io, (aq/> 312(aq) + 3H20(1) Next, the iodine is then free to react with the starch to form the familiar dark-blue colored complex. l2(aq)+ starch -> dark-blue colored complex Step 1: Prepare the following mixtures and calculate the total number of moles of iodate ion present in each 0.2 M KIO, solution for each mixture. (1000 mL = 1L) 0.1 M Sulfuric Distilled 0.20 M KIO Solution 5 ml Solution A Mixture I Mixture II Mixture Ill Mixture IV Acid 0 mL 0.5 mL 0 mL 0.5 mL water 15 mL 14.5mL 10 mL 9.5 mL 5 mL 10 mL 10 mL Step 2: Transfer 20 mL of each mixture to a flask, add 20 ml of Solution B, and start the timer. Record the time when the blue color appears. You may swirl the flask slowly if the color of the solution does not look same everywhere. Note that the total volume of the solution is 40 mL. You may repeat the experiment if you want. Record the time for reaction (seconds) Trial 1 Optional) Trial 2 Mixture l Solution B Mixture Il + Solution B :015 Mixture III + Solution B | . Mixture IV+ Solution 8 .03 Calculations: . (6pt) Calculate the total number of moles of iodate ion present in each 0.2 M KIOs solution the molarity of iodate ion in 40 mL of solution for each mixture. (1000 ml- i) . Calculate Solution A Volume of 0.2 M Moles of iodate ion Molarity of iodate ion irn KIO3 40 mL solution. Mixture I 5 mL Mixture II 5 mL Mixture 111 10 mL Mixture IV 10 mL 11. (8pt) Calculate Rate for each mixture Time (s) Rate Mixture I + Solution EB Mixture II+ Solution B | Mixture III + Solution B 0:oSs Mixture IV+ Solution B _Ill. (2 pts) Compare your rate results in mixtures Ill nd IV. What is the effect of adding sulfuric acid on the rate What do you think is the purpose of adding acid to this reaction?

Answer 1

KIO3 (aq) ---> K+ (aq) + IO3 - ( aq)

Hence, no. of moles of KIO3 = no. of moles of K+ = no. of moles of IO3- (iodate ion )

Molarity of KIO3 = 0.2 M

For mixture 1 ,

Volume of KIO3 = 5 ml = 0.005 L

Volume of H2SO4 = 0 ml

No. of moles of IO3- = molarity of KIO3 * volume of KIO3

= 0.2 * 0.005 = 0.001 mol ( answer)

In 40 ml solution , V = 40 ml = 0.040 L

Molarity = no. of moles / volume = 0.001 / 0.040 = 0.025 M ( answer )

For mixture 2,

Volume of 0.2 M KIO3 = 5 ml

Volume of 0.1 M sulphuric acid = 0.5 ml = 0.0005 L

Moles of sulphuric acid = 0.1 * 0.0005 = 0.00005 mol

H2SO4 (aq) ---> 2H+ (aq) + SO4- ( aq)

Hence no. of moles of H+ = 2* 0.00005 = 0.00010 mol

According to the given reaction between acid and KIO3 ,

6 moles of H+ react with 1 mol of IO3- completely

Hence, 0.00010 moles of H+ react with IO3- = 0.00010 / 6 = 0.0000166 moles

Hence no. of moles of IO3- left in solution = 0.001 - 0.0000166 = 0.000983 mol ( answer )

Molarity of 40 ml solution = 0.000983 / 0.040 = 0.02458 M ( answer )

Similarly, for mixture 3

Moles of KIO3 = 0.2 * 0.010 = 0.002 mol

Moles of H2SO4 = 0

Hence, moles of KIO3 = moles of IO3- = 0.002 mol ( answer )

Molarity of 40 ml solution

= 0.002 / 0.040 = 0.05 M ( answer )

Similarly for mixture 4,

Moles of KIO3 = 0.002 mol

Volume of H2SO4 = 0.5 ml

Molarity of H2SO4 = 0.1 M

Moles of H2SO4 = 0.1* 0.0005 = 0.00005 mol

Hence, no. of moles of IO3- reacted =2* 0.00005 / 6 = 0.0000166 mol ( similar to mixture 2 )

Hence moles of IO3- left = 0.002 - 0.0000166 = 0.0019834 mol ( answer )

Molarity of 40 ml solution = 0.0019834 / 0.040 = 0.0495 M ( answer )

I hope it helps. For further clarification please ask in comments.