i need help on both LUIUS. Question Completion Status: QUESTION 21 A 35.0 mL sample of...
A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH after the addition of 13.0 mL of NaOH? The Ka of benzoic acid is 6.3x10-5.
8) A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH at the eq. point? pka= 4.20
Remaining met hour, 15 minutes 44 seconds Question completion Status QUESTION 21 What is the pH of a solution prepared by mang 50.00L of 0.10 M and that the Ka NH -5.6 x 10-10 OR 9.16 11.13 QUESTION 22 What is the strongest monopotic acid of the following set if all the acid hydrofluoric acid with 3.5 x 10-4 Os benzoic acid with 6510-5 O cacetic acid with Ka - 1.8 x 10-5 lypochlorous acid with Ka - 3. 510-8...
this question. When a 21.0 mL sample of a 0.411 M aqueous hydrocyanic acid solution is titrated with a 0.300 M aqueous barium hydroxide solution, what is the pH after 21.6 mL of barium hydroxide have been added? PH Submit Answer Retry Entire Group 5 more group attempts remaining What is the pH at the equivalence point in the titration of a 23.3 ml sample of a 0.347 M aqueous hypochlorous acid solution with a 0.395 M aqueous barium hydroxide...
A 35.0-ml sample of 0 150 Macetic acid (CH,COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added. Part A O mL Express your answer using two decimal places. KALOOB ? pH Submit Request Answer Part B 17.5 ml. Express your answer using two decimal places EVO AZOO pl Subrni an
A 25.0 mL sample of 0.150 M formic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of formic acid is 1.8 ⋅ 10-4.
A 25.0 mL sample of 0.150 M chloroacetic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The of chloroacetic acid is 1.4 ×10-3.
JUL &coursejd_40296_1&new_attempt=1&content_ide_1144581.18 Question Completion Status: 35.0 mL of 0.10M propanoic acid CH3CH2COOH (K 1.3 x 10") is added to a 100.0 mL volumetric flask. Also added to the flask is 25.0 mL of 0.10M NaOH and water to fill the flask to the 100.0 ml line. How many moles of conjugate base would be formed? QUESTION 11 35.0 mL of 0.10M propanoic acid CH3CH2COOH (Ka - 1.3 x 10 ) is added to a 100.0 ml volumetric flask. Also added...
W ackboard Remaining Time: 23 minutes, 58 seconds. Question Completion Status: QUESTION 15 Which of the following is FALSE for a 0.150 M solution of a weak base with a kb - 1.40x10? The equilibrium concentration of hydroxide ions is 4.58x10-3M The solution would have a pH = 11.66. The equilibrium concentration of the conjugate acid would be 9.16x103 M. The conjugate acid would have a Ka - 7.14x10-11 The equilibrium concentration of the base would be 0.145 M. QUESTION...
Question 2 15 P Weak Acid Titration Calculations. A 25.0 mL aliquot of 0.50 M propanoic acid (CH3CH2O2H) is placed in a 250 mL Erlenmeyer flask and titrated with a standardized solution of 0.25 M NaOH. The pKof propanoic acid is 4.88. Calculate the pH of the solution after the addition of the following volumes of acid. Neglect activity. (15 points) a) 0.00 mL of NaOH added: 2.59 b) 50.00 mL of NaOH added: 8.49 c) 75.00 mL of NaOH...