Determine the resulting pH when 0.040 mol of solid NaOH is added to a 200.0 mL buffer containing 0.100 mol C6HşNH3CI and 0.500 M C6H5NH2. The value of Kb for C6H5NH2 is 4.3 × 10-1º. PREV 1 2 3 4 NEXT > Based on your ICE table and definition of Kb, set up the expression for Kb in order to determine the unknown. Do not combine or simplify terms. = 4.3 x 10-10 5 RESET [0] [0.500] [0.100] [0.040] [0.700] [0.300] [x] [2x] [0.500 + x] [0.500 - x] [0.100 + x] [0.100 - x] [0.040 + x] [0.040 - x] [0.700 + x] [0.700 - x] [0.300 + x] [0.300 - x]

Determine the mass of solid NaCH,COO that must be dissolved in an existing 500.0 mL solution of 0.200 M CH;COOH to form a buffer with a pH equal to 5.00. The value of Ka for CH,COOH is 1.8 x 10. NEXT > Let x represent the original concentration of CH;COO in the water. Based on the given values, set up the ICE table in order to determine the unknown. CH;COOH(aq) + H2O(1) H,O (aq) CH;COO (aq) Initial (M) Change (M) Equilibrium (M)

A 1.48 L buffer solution consists of 0.128 M butanoic acid and 0.338 M sodium butanoate. Calculate the pH of the solution following the addition of 0.080 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The K, of butanoic acid is 1.52 x 10-5. pH =

Calculate the pH change that takes place when a 150-mL portion of a) 0.0800 M NaOH and b) 0.0800 M HCl is added to 500 mL of the buffer solution containing 0.400 M NH3 and 0.500 M NH4Cl. Ka = 5.70 x 10-10 Calculate also the buffer capacity in a) and b).

A 20.3 mL sample of 0.370 M diethylamine, (C₂H5)2NH, is titrated with 0.366 M hydrochloric acid. At the equivalence point, the pH is|| Use the Tables link in the References for any equilibrium constants that are required.

10)

A 20.5 mL sample of 0.318 M methylamine, CH3NH₂, is titrated with 0.237 M nitric acid. At the equivalence point, the pH is Use the Tables link in the References for any equilibrium constants that are required.

Determine the [H3O+H3O+] of the eight buffer solutions made by mixing 1.10 M acetic acid and 0.900 M sodium acetate solutions according to the following volumes. The pKa for acetic acid, CH3CO2HCH3CO2H (HAc), is 4.745. You will also need this information in your post-lab. It is highly suggested you write these calculations down (both unrounded and rounded values) to use later. Solution 1.10 M Acetic Acid (mL) 0.900 M Sodium acetate (mL) [H3O+H3O+] (M) 1 8.00 2.00 Solution 1 Concentration 2 7.00 3.00 Solution 2 Concentration 3 6.00 4.00 Solution 3 Concentration 4 5.00 5.00 Solution 4 Concentration 5 4.00 6.00 Solution 5 Concentration 6 3.00 7.00 Solution 6 Concentration 7 2.00 8.00 Solution 7 Concentration 8 1.00 9.00 Solution 8 Concentration Solution 1 Concentration:   M Solution 2 Concentration:   M Solution 3 Concentration:   M Solution 4 Concentration:   M…

A 27.4 mL sample of 0.273 M trimethylamine, (CH3)3N, is titrated with 0.320 M hydroiodic acid. At the equivalence point, the pH is Use the Tables link in the References for any equilibrium constants that are required.

Ksp of Lead iodide (Pbl2) is 1.4 x 10 0. Calculate the molar solubility of this compound. 4.5 x 10-6M O 3.5 x 10-5M O2.5x 10-4M O 15 x 10-3M