Find the pH of these compounds. use 5% approximation where nescessary. say each compund has m...
Calculate Ka for 1 M NaCH3COO pH = 9.37 H+ = 4.203 X 10^-10 OH- = 2.402 X 10^-5 CH3COO- = 1.00 X 10^0 CH3COOH = 2.402 X 10^-5 ________________ Calculate Ka using the formula Kw = Ka X Kb and the appropriate equilibrium constant for 1 M NaCH3COO. NaCH3COO ---> Na+ (neutral) + CH3COO-. Is the conjugate base for CH3COOH. CH3COOH Ka1 = 1.8 X 10^-5 ______________________ Calculate Ka for 1 M NH4Cl pH = 4.26 H+ = 2.336...
1.) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 x 10-4) by 0.100 M HCl 200.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M HCl by 0.100 M NaOH...
1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M KOH by 0.100 M HCl 200.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCl 100.0 mL of 0.100 M HI by 0.100 M NaOH 100.0 mL of 0.100 M...
Which one of the following solutions has the lowest pH? A. a 0.1 M solution of formic acid, HCOOH (Ka = 1.8 x 10-4) B. a 0.1 M solution of ammonium chloride, NH4Cl(s) (Ka (NH4+) = 5.6 x 10-10) C. a 0.01 M solution of acetic acid, CH3COOH (Ka = 1.8 x 10-5) D. a 0.01 M solution of formic acid, HCOOH (Ka = 1.8 x 10-4) E. a 0.1 M solution of acetic acid, CH3COOH (Ka = 1.8 x...
A. Match each type of titration to its pH at the equivalence point. Weak acid, strong base Strong acid, strong base Weak base, strong acid pH less than 7 pH equal to 7 pH greater than 7 B. A 56.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 28.0 mL of KOH. C. Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8 x 10^-5) with 0.20 M HNO3....
Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH). 1 2 3 4 5 100.0 mL of 0.100 M HNO2 (Ka = 4.0 x 10-4) by 0.100 M NaOH 1 2 3 4 5 100.0 mL of 0.100 M HOCl (Ka = 3.5 x 10-8) by 0.100 M NaOH 1 2 3 4 5 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 x 10-4) by 0.100...
Titrations Part A A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 19.0 mL of HNO3. Express your answer numerically. Part B A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of NaOH. Express your answer numerically.
1. Calculate the equilibrium concentrations of all species present and the pH of a solution obtained by adding 0.100 moles of solid NaOH to 1.00 L of 15.0 M NH3. Kb = 1.8 × 10–5 2. One mole of a weak acid HA was dissolved in 2.0 L of water. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate the Ka for this weak acid. 3. Calculate the pH of a...
Need Help on 1-6 please
1. Mercury ions exit as the mercurous dimer Hg22+ and Hgt. If Hg22+ is in contact with liquid mercury, Hg (1), the following equilibrium is established: Hgt(aq) → Hg22+(aq) + Hg(1) K= 2.24x10-5 If a 0.15 M solution of Hgt is put in contact with liquid mercury, what will be the resulting equilibrium concentration of Hg22+? Show work a. 3.36 x 10-6 M b. 1.83 x 10-3 M c. 1.14 x 10-2 M d. 0.83...
Part C A 75.0-mL volume of 0.200 M NH3 (Kb = 1.8 x 10-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 15.0 mL of HNO3 . Express your answer numerically. Part D A 52.0-mL volume of 0.35 M CH3COOH (Ka = 1.8 x 10-5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 19.0 mL of NaOH. Express your answer numerically.