2.500g sample of a mixture of sodium carbonate and sodium chloride is dissolved in 25.00ml of 0.798M HCl. Some acid remains after the treatment of the sample. If 28.7 ml of 0.0108M NaOH were required to titrate the excess hydrochloric acid how many moles of sodium carbonate were present in the original sample?
How do i go about working this problem?
I know it has to do somthing with changing the Molarity into moles, but what do i do to get the original sample of moles present.
molesHCl=volueHCL*molarityHCl
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To solve this problem, you can follow these steps:
1. Calculate the number of moles of HCl used to titrate the mixture of sodium carbonate and sodium chloride:
moles of HCl = concentration of HCl × volume of HCl used (in liters)
moles of HCl = 0.798 M × 0.02500 L
Note: The given volume of HCl is converted to liters by dividing by 1000.
2. Calculate the number of moles of HCl that remained:
moles of HCl remaining = moles of HCl initially added - moles of HCl used
Note: The moles of HCl used are obtained from step 1.
3. Calculate the number of moles of NaOH used to titrate the remaining HCl:
moles of NaOH = concentration of NaOH × volume of NaOH used (in liters)
moles of NaOH = 0.0108 M × 0.0287 L
Note: The given volume of NaOH is converted to liters by dividing by 1000.
4. Since the equation between HCl and NaOH is 1:1, the moles of NaOH used are equal to the moles of HCl remaining.
5. Since the equation between HCl and Na2CO3 (sodium carbonate) is 2:1, the moles of HCl remaining are equal to twice the moles of Na2CO3 present in the original sample.
6. Finally, calculate the moles of Na2CO3:
moles of Na2CO3 = moles of HCl remaining / 2
By following these steps, you can determine the number of moles of sodium carbonate present in the original sample.
To solve this problem, you need to use the concept of acid-base titration. The key steps involved are:
1. Convert the volume (in mL) of the HCl solution to liters by dividing it by 1000.
- In this case, the volume of HCl solution is given as 25.00 mL. Convert it to liters: 25.00 mL ÷ 1000 = 0.02500 L.
2. Calculate the number of moles of HCl present in the solution by multiplying the volume (in L) by the molarity.
- The molarity of HCl is given as 0.798 M. Multiply it by the volume in liters:
Moles of HCl = 0.02500 L × 0.798 M = 0.01995 moles.
3. Determine the number of moles of NaOH used to titrate the excess HCl by multiplying the volume (in L) by the molarity.
- The volume of NaOH solution used is given as 28.7 mL. Convert it to liters: 28.7 mL ÷ 1000 = 0.0287 L.
- The molarity of NaOH is given as 0.0108 M. Multiply it by the volume in liters:
Moles of NaOH = 0.0287 L × 0.0108 M = 0.000310 mol.
4. Calculate the moles of excess HCl remaining after the treatment.
- Since NaOH reacts with HCl in a 1:1 ratio, the moles of HCl remaining equal the moles of NaOH used:
Moles of excess HCl = 0.000310 mol.
5. Determine the moles of HCl that reacted with Na2CO3, which will be equal to the moles of Na2CO3 present in the original sample.
- From the balanced equation, we know that 2 moles of HCl react with 1 mole of Na2CO3. Therefore, the moles of Na2CO3 can be calculated using the following ratio:
Moles of Na2CO3 = (0.01995 mol - 0.000310 mol) ÷ 2 = 0.00982 mol.
Therefore, the original sample contained 0.00982 moles of sodium carbonate (Na2CO3).