2. What is the resulting pH when 0.005 moles of KOH is added to 0.100 L of a buffer solution that is 0.100 M in H2PO4- and 0.100 M HPO4 2- and the Ka2= 6.2 x 10 -8.
3. Calculate the molar solubility of thallium chloride in 0.20M NaCl at 25 degrees Celcius. Ksp for TICl is 1.7 x 10 -4.
4. What volume of 0.500 M HNO3 is needed to titrate 60.00 mL of 0.500 M Ca(OH)2 to the equivalence point?
5. What is the chromium ion concentration for a saturated solution of Cr(OH)3 if the Ksp for Cr(OH)3 is 6.7 x 10 -31?
6. What is the molar solubility of Mg(OH)2 in a basic solution with a pH of 12.00? Ksp for Mg(OH)2 is 5.6 x 10-12.
7. What is the pH of a solution made by mixing 15.00mL of 0.10M acetic acid with 15.00mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. Ka=1.8 x 10-5 for CH3CO2H.
8. Formic acid (HCO2H, Ka=1.8 x 10 raised to the -4) is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00mL of the formic acid solution requires 49.80mL of 0.0567 M NaOH to reach the equivalence point?
1. What is the change in pH when 0.005 moles of HCl is added to 0.100 L of a buffer solution that is 0.100 M in CH3CO2H and 0.100 M NaCH3CO2? The Ka for acetic acid is 1.8 x 10 -5.
A buffer solution is one in which the pH of the solution is "resistant" to small additions of either a strong acid or strong base. Buffers usually consist of a weak acid and its conjugate base, in relatively equal and "large" quantities. Calculations are based on the equation for the ionization of the weak acid in water forming the hydronium ion and the conjugate base of the acid. "HA" represents any weak acid and "A-" represents the conjugate base.
HA(aq) + H2O(l) --> H3O+(aq) + A-(aq)
When a strong acid is added to a buffer solution the conjugate base present in the buffer consumes the hydronium ion converting it into water and the weak acid of the conjugate base.
A-(aq) + HCl+(aq) --> H2O(l) + HA(aq)
So your first step is to write the equation for the ionization of acetic acid in water and the related Ka expression rearranged to solve for the hydronium ion concentration.
CH3COOH(aq) + H2O(l) --> H3O+(aq) + CH3COO-(aq)
[H3O+] = Ka[CH3COOH][CH3COO-]
Next, make and ICE chart, which I have attached. Let "x" represent the hydronium ion concentration once equilibrium has been re-established. We will assume that all of the added acid is consumed.
Thirdly, substitute into the Ka expression and solve for the hydronium ion concentration. Convert ...
Organic chemistry problems involving buffers, pH calculations, and equivalence points.