concentration of excess NaOH(Titrant) = 1*0.25/(25+20+1) = 0.00543 M
pH = 14 -(- log(OH-))
= 14 - (-log(0.00543))
= 11.74
answer: C
43. To reach the stoichiometric point in the titration of 25.0 mL of a 0.200 M...
You are charged to perform a titration with 20 mL of 1.0 M hydrazine (KB = 1.7x10-6) as analyte and 0.5 M HBr as titrant. What is the initial pH of the solution? pH = At what volume would you reach the stoichiometric point? stoichiometric point = mL At what volume would pH = pka? pH = pka @ ml What is the pH at the stoichiometric point? pH @ stoichiometric point =
IUS. A 21.54-mL volume of 0.130 M NaOH is required to reach the stoichiometric point for the titration of 25.00 mL of a 0.112 M HCl solution. Would the titration of 25.00 mL of a 0.112 M CH3COOH solution require more, less, or the same volume of the 0.130 M NaOH solution? Explain. 6 - -
23. What is the pH at the stoichiometric point for the titration of 50.00 mL of 0.100 M of HNO2(ag) with 0.200 M NaOH(ag)? The value of Ka for HNO2 is 1.0 x 10-7
ASAP please
Consider the titration of 50.0 mL of 0.200 M hypochlorous acid HCIO (Ka = 3.5 x 10-8) with 0.250 M NaOH. 5- How many milliliters of NaOH are required to reach the equivalence point? a) b) Calculate the pH before titration c) Calculate the pH after adding 40.0 mL. of NaOH d) Calculate the pOH after adding 20.0 ml, of NaOH
Assume a titration with 0.100 M NaOH titrant and 25.00 mL of a 0.0800 M CH3COOH analyte. How many mL of NaOH is required to reach the equivalence point? Assume a titration with 0.100 M NaOH titrant and 25.00 mL of a 0.0800 M CH3COOH analyte. What will the initial pH of the analyte be if 0.00 mL of NaOH is added?
You perform a titration using 25.0 mL of a 0.20 M propionic acid (HC3H5O2) as your analyte and 0.25 M NaOH as your titrant. Assume a new 25.0 mL of analyte is used for each step, determine the pH when the following volumes of titrant are used. (Propionic acid Ka= 1.32 x 10^-5) a. 15.0 mL titrant added. b. 25.0 mL titrant added.
Consider the titration of 25.00 mL of 0.200 M methyl amine (CH3NH2). The titrant is 0.120 M HCl. Calculate each of the following: a. the volume of added acid required to reach the equivalence point. b. the pH at ½ the HCl needed to reach the equivalence point c. the pH at the equivalence point.
By titration, it is found that 37.5 mL of 0.200 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.
42. What is pH at the stoichiometric point of a titration of 25.0 mL of a 0.100 Msolution of methylamine (pKb 4.20) with a 0.125 M solution of HCI? A. 5.40 B. 8.47 C. 7.00 D. 8.60 5.53
24. A 25.0 mL volume of a 0.200 M N,H& solution (K 1.70x10 titrated to the equivalence point with 0.100 M HCl. What is the pH of this solution at the equivalence point? The titration is a. 4.70 b. 8.23 c. 7.00 d. 9.30
24. A 25.0 mL volume of a 0.200 M N,H& solution (K 1.70x10 titrated to the equivalence point with 0.100 M HCl. What is the pH of this solution at the equivalence point? The titration is...