A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium ca
ID: 776131 • Letter: A
Question
A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.110g of this sample requires 23.98 mL of 0.100 MHCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?Carbonic acid (H2CO3) is a weak diprotic acid with Ka1=4.4310?7 and Ka2=4.7310?11. When sodium bicarbonate (NaHCO3) is titrated with hydrochloric acid (HCl), it acts as a weak base according to the equation
NaHCO3(aq)+HCl(aq)?H2CO3(aq)+NaCl(aq)
Suitable indicators are those that change color within the pH range for the equivalence point of a specific titration. The expected pH at the equivalence point can be calculated using pKa values. Suitable indicators for use in titrating carbonic acid or carbonate A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.110g of this sampleequires 23.98 mL of 0.100 MHCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?solutions are methyl orange and phenolphthalein.
A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.110g of this sample requires 23.98 mL of 0.100 MHCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?
Explanation / Answer
First we will find out how many moles of NaHCO3 we have by calculating the molecular mass:
1Na(23.0) + 1H(1.008) + 1C(12.0) + 3O(16.00) =84.008 g/mole
1.10g(1mole/84.008g)=0.0131 moles of NaHCO3
Let v = volume of HCl
Assume one mole of HCl neutralizes one mole of NaHCO3.
v(0.200moles/L)=0.0131moles
v=0.0131moles/0.200moles/L
v=0.0655L or 65.5mL
3. Phenolphthalein endpoint is for the strong base NaOH.
23.98mL(1L/1000mL)(0.100moles/L)=0.002 moles
molecular mass NaOH = 23.0 + 16.0 + 1.008 =40.008g/mole
mass NaOH = 0.002398moles(40.008g/mole) = 0.0959g
The methyl orange endpoint is for the Na2CO3.
0.700mL(1L/1000mL)(0.100moles/L) = 0.000
molecular mass is 2(23.0)+12.0 + 3(16.0) =106.0g/mole
mass Na2CO3= 0.0000700moles(106.0g/mole) = 0.00742g
Total mass is 0.00742+0.0959=0.1033g
%Na2CO3=(0.00742g/0.1033g)100%= 7.182 %