Question

Carbonic acid (H2CO3) is a weak diprotic acid with Ka1=4.43×10−7 and Ka2=4.73×10−11. 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 solutions are methyl orange and phenolphthalein.

Part A What volume of 0.130 M HCl is required for the complete neutralization of 2.00 g of NaHCO3(sodium bicarbonate)? What volume of 0.170 M HCl is required for the complete neutralization of 1.50 g of Na2CO3(sodium carbonate)? A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.190 g 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?

Answer 1

What volume of 0.130 M HCl is required for the complete neutralization of 2.00 g of NaHCO3(sodium bicarbonate)?

NaHCO3(aq) + HCl (aq)= NaCl(aq)+ H2CO3(aq)

First calculate the moles of NaHCO3 in 2.00 g

Number of moles = amount in g/ molar mass

= 2.00 g / 84.007 g/ mole

= 0.024 moles

NaHCO3 and HCl Reacted in 1:1 therefore the moles of HCl are 0.024 moles

Molarity = number of moles / volume in L

Volume in L = number of moles / molarit

= 0.024 Moles / 0.130 M or moles / L

= 0.1846 L

= 185.6 L

What volume of 0.170 M HCl is required for the complete neutralization of 1.50 g of Na2CO3(sodium carbonate)?

NaHCO3(aq) + HCl (aq)= NaCl(aq)+ H2CO3(aq)

First calculate the moles of NaHCO3 in 2.00 g

Number of moles = amount in g/ molar mass

= 1.50 g / 84.007 g/ mole

= 0.018 moles

NaHCO3 and HCl Reacted in 1:1 therefore the moles of HCl are 0.018 moles

Molarity = number of moles / volume in L

Volume in L = number of moles / molarit

= 0.018 Moles / 0.170 M or moles / L

= 0.1050 L

= 105.0 L

A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.190 g 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?

Phenolphthalein endpoint is for the strong base NaOH.

Number of moles = volume in L* Molarity

= 23.98mL(1L/1000mL)(0.100moles/L)

=0.002... moles
molecular mass NaOH=40.008g/mole

Mass = molar mass * number of moles
mass NaOH = 0.002398moles(40.008g/mole)=0.0959g

The methyl orange endpoint is for the Na2CO3.

Number of moles = volume in L* Molarity
=0.700mL(1L/1000mL)(0.100moles/L)

=7.0*10^-5 ..
molecular mass of Na2CO3.= 106.0g/mole

mass Na2CO3= 0.0000700moles(106.0g/mole)

=0.00742g

Total mass is 0.00742+0.0959=0.1033g

%Na2CO3=mass of Na2CO3 / total mass *

=(0.00742g/0.1033g)100

=7.18%