A 1.00 L solution contains 0.352 M HA (Ka = 4.66⋅10−64.66⋅10-6) and 0.758 M NaA.
What is the pH of this solution? | |
What is the pH if 0.310 L of 0.138 M KOH is added to the solution be? | |
What is the pH if 0.620 L of 0.138 M KOH is added to the solution be? |
Given solution contain weak acid HA and its salt NaA. Therefore this solution is a buffer solution. We can use Henderson's equation to calculate pH of buffer solution.
We have Henderson's equation, pH = pKa + log [ Salt ] / [ acid ]
pH = - log Ka + log [ NaA ] / [ HA ]
pH = - log 4.66 10 -06 + log 0.758 / 0.352
pH = 5.332 + log 0.758 / 0.352
pH = 5.332 + 0.333
pH = 5.665
Question 2
Calculation of moles of NaA, HA and KOH.
We have relation, Molarity = No. of moles of solute / Volume of solution in L
No. of moles of solute = Molarity volume of solution in L
No. of moles of HA = 0.352 mol / L 1.00 L = 0.352 mol
No. of moles of NaA = 0.758 mol / L 1.00 L = 0.758 mol
No. of moles of KOH = 0.138 mol / L 0.310 L = 0.04278 mol
Consider reaction of KOH with buffer solution.
HA + OH - A - + H2O
Let's use ICE table.
Moles | HA | OH - | A - |
I | 0.352 | 0.04278 | 0.758 |
C | -0.04278 | -0.04278 | +0.04278 |
E | 0.30922 | 0.00000 | 0.80078 |
Volume of buffer solution after addition of KOH = 1.00 L + 0.310 L = 1.310 L
We have equation, pH = - log Ka + log [ NaA ] / [ HA ]
pH = - log 4.66 10 -06 + log ( 0.80078 mol / 1.310 L ) / ( 0.30922 mol / 1.310 L )
pH = 5.332 + 0.4132
pH =5.745
Question 3
Calculation of moles of NaA, HA and KOH.
We have relation, Molarity = No. of moles of solute / Volume of solution in L
No. of moles of solute = Molarity volume of solution in L
No. of moles of HA = 0.352 mol / L 1.00 L = 0.352 mol
No. of moles of NaA = 0.758 mol / L 1.00 L = 0.758 mol
No. of moles of KOH = 0.138 mol / L 0.620 L = 0.08556 mol
Consider reaction of KOH with buffer solution.
HA + OH - A - + H2O
Let's use ICE table.
Moles | HA | OH - | A - |
I | 0.352 | 0.08556 | 0.758 |
C | -0.08556 | -0.08556 | +0.08556 |
E | 0.26644 | 0.00000 | 0.84356 |
Volume of buffer solution after addition of KOH = 1.00 L + 0.620 L = 1.620 L
We have equation, pH = - log Ka + log [ NaA ] / [ HA ]
pH = - log 4.66 10 -06 + log ( 0.84356 mol / 1.620 L ) / ( 0.26644 mol / 1.620 L )
pH = 5.332 + 0.5005
pH =5.832
A 1.00 L solution contains 0.352 M HA (Ka = 4.66⋅10−64.66⋅10-6) and 0.758 M NaA. What...
A solution contains 0.352 M HA (Ka = 4.66⋅10−64.66⋅10-6) and 0.758 M NaA. What is the pH of this solution? What is the pH of this solution after 0.138 mol of HCl are added to 1.00 L of this solution? What is the pH of this solution after 0.276 mol of HCl are added to 1.00 L of this solution?
A 1.00 L solution contains 0.311 M HA (K3 = 3.04 - 10 ) and 0.796 M NaA. What is the pH of this solution? What is the pH if 0.318 L of 0.158 M KOH is added to the solution be? What is the pH if 0.636 L of 0.158 M KOH is added to the solution be?
A 1.00 L solution contains 0.326 M HA (Kg = 5.09 · 10-7. and 0.796 M NaA. What is the pH of this solution? What is the pH if 0.325 L of 0.219 M KOH is added to the solution be? What is the pH if 0.650 L of 0.219 M KOH is added to the solution be?
A solution contains 0.250 M HA (К.-1.0 added to 1.00 L of this solution? 3. 10" and 0.45 M NaA. what is the pH after 0.30 mole of HC1 is a. 0.52 b. 7.44 c. 5.44 d. 2.10 e. 8.56
a) A solution contains 0.34 M of a weak acid HA (Ka = 2.0 x 10-7) and 0.17 M NaA. What is the pH after 0.05 M of HCl is added to this solution (assume no volume change) b) The pH of a solution containing 0.1 M of a weak acid HA is 6. Calculate Ka for this acid. Note the acids are unrelated for the two problem parts.
A solution contains 0.250 MHA(Ka-1.0 x 106and 0.45 M NaA. What is the pH after 0.30 mole of HCl is added to 1.00 L of this solution? a. 7.44 .2.10 98.56 8.5.44 e.0.52
Calculate the pH of a 0.250 M solution of a sodium salt, NaA. pKa HA = 4.66
A 0.25 M solution of a salt NaA has pH=8.93. What is the value of Ka for the parent acid HA?
1. Consider a 0.475 M NaA(aq) solution in which HA(aq) has a Ka of 1.950E-03 at 25°C. a. Identify the major species present in the NaA(aq) solution. Major species: b. Calculate the Kb of the conjugate base, A (aq). Kb = c. Calculate the pH of a 0.475M NaA(aq) solution.
Question 1 : HA is a weak acid. Its ionization constant, Ka, is 1.2 x 10-13. Calculate the pH of an aqueous solution with an initial NaA concentration of 0.075 M. Question 2 : We place 0.143 mol of a weak acid, HA, in enough water to produce 1.00 L of solution. The final pH of the solution is 1.28 . Calculate the ionization contant, Ka, of HA. Question 3 : We place 0.661 mol of a weak acid, HA,...