Zane titrated 25.0 mL of 0.40 M propioic acid with 0.20 M NaOH. What volume of NaOH is needed to reach the equivalence point? KaHA = 1.34 x 10-5 Кана Volume NaOH = [?] mL Volume NaOH (mL) Enter

Answer Questions #27 through #31 about the titration curves below (labeled a and b) for two different bases. The titrant acid used in both cases was 0.100 M HCl and the volume of the base solutions in both cases was 75.0 mL mL. 14 14 12 12 - 10- 10 8- 6. 4. 2- 2- 20 40 60 80 100 20 40 60 80 100 Volume of acid added (mL) (a) (b) Volume of acid added (mL) 2014 Pearson Education Inc P Type here to search 11:17 AM (? 3/18/2021 -> ort sc 80 6. ece num ER/TY || на 0 04 2C Hd

In a titration of 25.0 mL of 0.10 M phosphoric acid (a triprotic weak acid) with 0.20 M NaOH, what species are present after the addition of 20.0 mL of NaOH?What species would be present if we only added 10 mL NaOH?

1. How do you prepare a 100mL of 0.1 M phosphate buffer?To make 100 mL of 0.1 M phosphate buffer: Calculate the amount of sodium phosphate needed. moles of sodium phosphate = (0.1 mol/L) x (0.1 L) = 0.01 moles mass of sodium phosphate = moles x molar mass = 0.01 moles x 142 g/mol = 1.42 g Dissolve 1.42 g of sodium phosphate in distilled water in a 100 mL volumetric flask. Adjust the pH of the solution to the desired value (usually around 7.4) using a strong acid or strong base. Bring the solution to the final volume (100 mL) with distilled water.   2. From the anterior buffer, how do you make 100mL of 0.05 M? To make 100 mL of 0.05 M phosphate buffer from the 0.1 M stock solution: Calculate the amount of the 0.1 M phosphate buffer needed. moles of phosphate buffer = (0.05 mol/L) x (0.1 L) = 0.005 moles Calculate the volume of the 0.1 M phosphate buffer needed. moles = concentration x volume (in liters) volume = moles / concentration = 0.005 moles / 0.1 mol/L = 0.05 L or 50 mL Measure…

Select the answer that best describes the experimental setup of the titration curve shown in the image below 12 Hd H] 8 4 0 10 volume of titrant (mL) 20 20 strong base (beaker) and strong acid (burette) weak base (beaker) and strong acid (burette) weak acid (beaker) and strong base (burette) strong acid (beaker) and strong base (burette)

Calculate the final [Ag1+] concentration in solution when 0.035 moles of AgNO3 are added to 500. mL of 0.21 M NaCN. Assume the volume remains constant. Ksp = 1.2 ✕ 10-16 for AgCN. (in M)

A 10.0mL solution of 0.720 M NH3 is titrated with a 0.240 M HCl solution. Calculate the pH after the following additions of the HCl solution: 0.00mL, 10.0mL, 30.0mL, and 40.0mL. Be sure your answer has the correct number of significant digits. 0.00 mL added, pH = 10.0 mL added, pH = 30.0 mL added, pH = 40.0 mL added, pH =

A 28.00 mLmL sample of an H2SO4H2SO4 solution of unknown concentration is titrated with a 0.1422 MM KOHKOH solution. A volume of 43.22 mLmL of KOHKOH was required to reach the equivalence point. What is the concentration of the unknown H2SO4H2SO4 solution?         Express your answer using four significant figures.

5 6 7 8 Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K, of hypobromous acid is 2.8 x 10¹⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide (LiOH). What is the initial solution pH (i.e., when 0.00 mL of titrant have been added)? (Two decimal places) Type your answer... Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K, of hypobromous acid is 2.8 x 10¹⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide (LiOH). What volume (in mL) of titrant will have been added at the half-equivalence point? (One decimal place) Type your answer... Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K₂ of hypobromous acid is 2.8 x 10.⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide…

Consider the titration of 80.0 mL of 0.100 M Ba (OH), by 0.400 M HCI. Calculate the pH of the resulting solution after the following volumes of HCl have been added. а. 0.0 mL pH = b. 20.0 mL pH = c. 30.0 mL pH = d. 40.0 ml pH = e. 90.0 ml pH =