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Title: A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of distilled water. Ph Post by: nappy110 on Oct 14, 2015 Hi guys, I was hoping I could get help with this problem please.
A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of distilled water. Phenolphthalein was added and 11.40 mL of 0.09942 M HCL was required to reach the first end point. Excess volume of HCL was added according to the lab procedure (if x is the amount of acid needed to complete the first titration, you will add a total of 2x + 10mL of acid to ensure excess). After boiling, the excess acid was titrated with 0.1020 M NaOH. It was found that 10.78 mL of the NaOH was required to reach the end point of this titration. Answer the following about this determination: 1. How many total mL of the 0.09942 M HCL will have been added if the lab procedure is followed? 2. Based on your answer to #1 above, how many total moles of HCL were added to the sample? 3. How many moles of NaOH were used in the back titration? 4. How many moles of the HCL in problem #2 were used up in reacting with sodium carbonate in the sample of unknown? 5. What was the % sodium carbonate in the original unknown sample? Title: A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of distilled water. Ph Post by: bio_man on Oct 14, 2015 1. the volume of HCl need is 2x+10 ml with x is the volume need for the first end point
2x+10 --> 2(11.40)+10 =32.80 ml HCl 2. nHCl= (32.80/1000)*0.09942 = 3.26*10^-3 mol HCl 3.nNaOH= (10.78/1000)*0.1020 = 1.1*10^-3 mol NaOH 4. nHCl react with the Na2CO3 is 3.26*10^-3 - 1.1*^-3 = 2.16*10^-3 mol HCl nHCl = n NaOH becuase the ratio is 1:1 5. 2nHCl = n Na2CO3 because the ratio is 2:1 2HCl + Na2CO3 --> 2NaCl + CO2 + H2O => nNa2CO3 = 2.16*10^-3 /2 = 1.08*10^-3 mol => m Na2CO3 = 1.08*10^-3 * (23+23+12+48) = 0.11448 g % Na2CO3 = 0.11448/0.4016 *100% = 28.5 % |