Consider the reaction that occurs on mixing 50.0 mL of 0.560 MNaHCO3 and 50.0 mL of 0.400 M NaOH at25oC.
(a) Write a balanced net ionic equation for thereaction.
(b) What is the pH of the resulting solution?
(c) How much heat (in joules) is liberated by thereaction?
(standard heats of formation are given in appendix B=> ÄHof(kJ/mol)NaHCO3= -950.8 and for NaOH = -425.6 )
(d) What is the final temperature of the solution to thenearest 0.1oC? You may assume that all the heatliberated is absorbed by the solution, the mass of the solution is100.0g, and its specific heat is 4.18 J/(g.C)
(a) The balanced net ionic equation for the reaction is:
2HCO3- + 2Na+ + 2OH- -> 2CO3^2- + 2Na+ + 2H2O
(b) To determine the pH of the resulting solution, we first need to determine the concentration of the resulting compounds.
The initial concentrations of NaHCO3 and NaOH are:
[NaHCO3] = 0.560 M
[NaOH] = 0.400 M
Since NaHCO3 is a weak acid and NaOH is a strong base, the reaction between them will result in the formation of the conjugate base of NaHCO3, which is HCO3-. HCO3- is a weak base.
The concentration of HCO3- in the resulting solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
For the reaction between NaHCO3 and NaOH, the pKa value can be taken as the pKa of carbonic acid (H2CO3), which is 6.35.
Using the Henderson-Hasselbalch equation, we can rearrange it to solve for [HCO3-]:
[HCO3-] = 10^(pH - pKa) * [NaHCO3]
Substituting the given values:
[HCO3-] = 10^(pH - 6.35) * 0.560 M
(c) To calculate the heat liberated by the reaction, we can use the following equation:
q = n * ΔH
where q is the heat (in joules), n is the number of moles of the limiting reactant, and ΔH is the heat of reaction.
First, we need to determine the limiting reactant.
NaHCO3 reacts with 2 equivalents of NaOH, so we need to find the number of moles of NaOH that reacts with the given amount of NaHCO3.
Number of moles of NaHCO3 = (volume of NaHCO3 solution in L) * (molarity of NaHCO3)
Number of moles of NaOH = (volume of NaOH solution in L) * (molarity of NaOH)
Since NaHCO3 and NaOH react in a 1:1 ratio, the limiting reactant is the one with the fewer moles.
Once we determine the limiting reactant, we can calculate the number of moles of the other reactant that reacts.
The heat liberated by the reaction can then be calculated using the equation above.
(d) To determine the final temperature of the solution, we can use the equation:
q = mcΔT
where q is the heat (in joules), m is the mass of the solution (in g), c is the specific heat (in J/(g°C)), and ΔT is the change in temperature (in °C).
Using the heat calculated in part (c), we can rearrange the equation to solve for ΔT:
ΔT = q / (mc)
Substituting the given values, we can calculate the final temperature of the solution.
(a) To write a balanced net ionic equation for the reaction between NaHCO3 and NaOH, we first need to write the balanced chemical equation.
The chemical equation for the reaction is:
NaHCO3 + NaOH -> Na2CO3 + H2O
To balance the chemical equation, we need to ensure that the number of atoms on both sides of the equation is equal.
NaHCO3 + NaOH -> Na2CO3 + H2O
The net ionic equation only includes the species that are directly involved in the reaction. In this case, the net ionic equation is:
HCO3- + OH- -> CO32- + H2O
(b) To calculate the pH of the resulting solution, we need to determine if the reaction is an acid-base reaction or not. In this case, the reaction between NaHCO3 and NaOH is an acid-base reaction. NaHCO3 is a weak acid (bicarbonate ion, HCO3-) while NaOH is a strong base (hydroxide ion, OH-).
When an acid and a base react, they neutralize each other to produce a salt and water. In this case, the salt formed is Na2CO3. Na2CO3 is a salt that dissociates into sodium cations (Na+) and carbonate anions (CO32-).
Since Na2CO3 is a salt, it dissociates completely in water, and the pH of the resulting solution depends on the hydrolysis of the water. The hydrolysis of water results in the formation of hydrogen ions (H+) and hydroxide ions (OH-).
In this case, the hydrolysis of water is expected to produce both H+ and OH-. However, since we have an excess of OH- from NaOH, the resulting solution will be basic. Therefore, the pH of the resulting solution will be greater than 7.
(c) To calculate the heat liberated by the reaction, we need to use the standard heats of formation for NaHCO3 and NaOH.
ΔHof(kJ/mol) NaHCO3 = -950.8 kJ/mol
ΔHof(kJ/mol) NaOH = -425.6 kJ/mol
The reaction is exothermic, meaning it releases heat. The molar ratios between NaHCO3 and NaOH in the balanced equation are 1:1.
To calculate the heat liberated, we can use the equation:
ΔHrxn = Σ(n * ΔHof(products)) - Σ(n * ΔHof(reactants))
Where:
ΔHrxn = heat of the reaction
n = stoichiometric coefficients
For this reaction:
ΔHrxn = (1 * ΔHof(Na2CO3) + 1 * ΔHof(H2O)) - (1 * ΔHof(NaHCO3) + 1 * ΔHof(NaOH))
(d) To calculate the final temperature of the solution, we need to use the equation:
q = m * C * ΔT
Where:
q = heat absorbed by the solution
m = mass of the solution
C = specific heat of the solution
ΔT = change in temperature
In this case, the heat absorbed by the solution is equal to the heat liberated by the reaction (from part (c)). We can then rearrange the equation to solve for the final temperature:
ΔT = q / (m * C)
Plug in the values of q (from part (c)), m = 100.0 g, and C = 4.18 J/(g.C) into the equation to find the final temperature of the solution.