Question

In: Chemistry

A titration involves adding a reactant of known quantity to a solution of an another reactant...

A titration involves adding a reactant of known quantity to a solution of an another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. The equation for the reaction of a generic weak acid HA with a strong base is HA(aq)+OH−(aq)→A−(aq)+H2O(l) A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 9.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH? More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 59.0 mL ?

Solutions

Expert Solution

HA(aq) +   OH−(aq)   -----------------> A−(aq)   + H2O(l)

9.00            1.00                                    0                0

8.00              0.00                                 1.00            

pKa = -log (5.61 x 10^-6) = 5.25

pH = pKa + log [salt / acid]

    = 5.25 + log [1.00 / 8.00]

pH = 4.35

b)

here salt is formed .

salt concentration = 9 / 59 = 0.1525 M

pH = 7 + 1/2 (pKa + log C)

    = 7 + 1/2 (5.25 + log 0.1525)

pH = 9.22


Related Solutions

Ten Titration of Weak Acid with Strong Base A titration involves adding a reactant of known...
Ten Titration of Weak Acid with Strong Base A titration involves adding a reactant of known quantity to a solution of another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. The equation for the reaction of a generic weak acid HA with a strong base is HA9(aq)+OH-(aq) -----> A-(aq)+H2O(l) A certain weak acid, HA, with a Ka value of 5.61x10-6, is titrated with NaOH Part A A solution is made by...
To convert from a given quantity of one reactant or product to the quantity of another...
To convert from a given quantity of one reactant or product to the quantity of another reactant or product: First, convert the given quantity to moles. Use molar masses to convert masses to moles, and use Avogadro's number (6.02×1023 particles per mole) to convert number of particles to moles. Next, convert moles of the given reactant or product to moles of the desired reactant or product using the coefficients of the balanced chemical equation. For example, in the chemical equation...
The iodine clock reaction involves reacting an aqueous solution of potassium iodate with another solution containing...
The iodine clock reaction involves reacting an aqueous solution of potassium iodate with another solution containing sulfuric acid, sodium bisulfite, and starch. The reaction begins colourless but at a certain point the free triiodide concentration builds up, complexing with the starch molecules and producing a dark blue to almost black colour almost instantaneously. Which of the following options will cause the blue colour to appear faster? Increasing pH Increasing bisulfite concentration Increasing starch concentration Decreasing temperature Increasing temperature
Outline of procedure: Dilute bleach solution A 1/10. Prepare analyte for titration by adding 10mL of...
Outline of procedure: Dilute bleach solution A 1/10. Prepare analyte for titration by adding 10mL of diluted bleach solution, indicator and 5mL of 2 M HCl. Titrate the analyte with 0.05524 M Na2S2O3. Average volume of Na2S2O3 added = 31.25 mL Question: Calculate the percent of sodium hypochlorite in the original bleach bottle using your averaged result. Show the calculation for solution A only. Which bleach solution is the more potent cleaning agent?
A precipitation titration was performed by adding 0.20 M Ag+ (from standard silver nitrate solution) from...
A precipitation titration was performed by adding 0.20 M Ag+ (from standard silver nitrate solution) from a buret into 40.0 mL mixture of 0.10 M I- (as NaI) and 0.10 M Cl- (as NaCl). Calculate the concentration of Ag+, I-, Cl-, Na+, and NO3- when (a) 0.0 mL, (b) 10.0 mL, (c) 20.0 mL, (d) 30.0 mL, (d) 40.0 mL, and (d) 50.0 mL of Ag+ were added. (Given Ksp AgI = 8.3 x 10-17 and Ksp AgCl = 1.8...
An indirect titration is a titration which requires the presence of another species to determine the...
An indirect titration is a titration which requires the presence of another species to determine the endpoint. When triiodide (I3-) is used to determine the concentration of formaldehyde in solution, a back titration is used. For what reason is a direct titration not used? What is the third species that is added to the titration? _________________________ Describe how the other two species present during the titration work in conjunction to allow the analyst to determine the formaldehyde concentration Distinguish between...
Converting between quantities To convert from a given quantity of one reactant or product to the...
Converting between quantities To convert from a given quantity of one reactant or product to the quantity of another reactant or product: First, convert the given quantity to moles. Use molar masses to convert masses to moles, and use Avogadro's number (6.02×1023particles per mole) to convert number of particles to moles. Next, convert moles of the given reactant or product to moles of the desired reactant or product using the coefficients of the balanced chemical equation. For example, in the...
1- In the titration between potassium iodate and the sodium thiosulfate solution, if the titration is...
1- In the titration between potassium iodate and the sodium thiosulfate solution, if the titration is not performed immediately after the addition of the sulfuric acid, how would this likely affect the calculated concentration of your diluted sodium thiosulfate solution? 2- Why did the titrations performed need to be started immediately after adding the sulfuric acid? 3- A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach solution...
A titration is performed by adding .435M KOH to 80ml of .205M HC3H5O2. A) Calculate the...
A titration is performed by adding .435M KOH to 80ml of .205M HC3H5O2. A) Calculate the pH before the addition of any KOH? B) Calculate the pH after the addition of 7.54ml of KOH? C) Calculate the volume of base needed to reach the equivalence point? D) Enter the pH of the solution at the equivalence point of the titration?
The amount of l3 (aq) in a solution can be determined by titration with a solution...
The amount of \(\mathrm{I}_{3}{ }^{-}(a q)\) in a solution can be determined by titration with a solution containing a known concentration of \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}(a q)\) (thiosulfate ion). The determination is based on the net ionic equation$$ \mathbf{2 S}_{2} \mathbf{O}_{3}^{2-}(a q)+\mathbf{I}_{3}^{-}(a q) \rightarrow \mathbf{s}_{4} \mathbf{O}_{6}^{2-}(a q)+3 \mathbf{I}^{-}(a q) $$Given that it requires \(34.3 \mathrm{~mL}\) of \(0.360 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)\) to titrate a \(30.0-\mathrm{mL}\) sample of \(\mathrm{I}_{3}^{-}(\mathrm{aq})\), calculate the molarity of \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) in the solution.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT