8. Construct a theoretical curve for the titration of HAC by sodium hydroxide (NaOH). The volume of acetic acid: in the beaker is 25 ml and its concentration is 0.02 M. The titrant is NaOH and its concentration is also 0.02 M. Calculate pH versus volume of NaOH, which will be added in increments of 1 mL. Construct this titration curve for 35 mL. Show only the calculations at VNaoH=0, 0 < VNaOH < Ve, VNaOH=Ve and VNaOH > Ve). 14 a) 금 7 equivalence point Vbase Hd 14 b) equivalence point (Ve) 7 pka buffering region Ve2 Ve Vbase

A chemist titrates 100.0 mL of a 0.0621 M hydrobromic acid (HBr) solution with 0.6950M KOH solution at 25 °C. Calculate the pH at equivalence. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = 0 X 5

A scientist is attempting to recreate this titration using the same concentration of hydrochloric acid, but decides to use a lower concentration of sodium hydroxide(titrant). The original concentration of sodium hydroxide is 0.1M. How will this affect the amount of titrant needed? Explain.Correctly predicts the relative amount of titrant needed and explains.

A chemist titrates 150.0 mL of a 0.8461M hydrobromic acid (HBr) solution with 0.2857M NAOH solution at 25 °C. Calculate the pH at equivalence. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added. PH = |

10 11 12 13 14 15 16 17 18 19 20 A chemist titrates 170.0 mL of a 0.3333 M nitric acid (HNO, solution with 0.1703M NaOH solution at 25 °C. Calculate the pH at equivalence. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the soluțion equals the initial volume plus the volume of NaOH solution added. pH = %3D Submi Continue MacBook Air

Determine the Ka value Solution 1 3.98 X 10-3 M Solution 2 6.03 X 10-4 M Solution 3 1.07 X 10-5 M Solution 4  1.05 X 10-5 M Solution 5  4.68 X 10-10 M

A chemist titrates 150.0 mL of a 0.6793 M potassium hydroxide (KOH) solution with 0.0962 M HNO, solution at 25 °C. Calculate the pH at equivalence. Round your answer to 2 decimal places Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO, solution added. pH = 0

Let HA represent a weak acid. A 10.0 mL sample of 0.600M HA was titrated with 0.300 M NaOH. The table below is some of the data for this titration. Volume NaOH added (mL) 5.00 10.00 15.00 ? ? pH 6.98 7.46 7.93 ? ? Calculate the following: a) The volume of NaOH needed to reach the equivalence point is mL. (Report value to the correct number of significant figures.) b) The Ka for HA is Report value using no superscripts. For example, Kw = 1.0 x 10-14 or 1.0e-14 (Report value to the correct number of significant figures.) c) The Kp for A is Report value using no superscripts. For example, Kw = 1.0 x 10-14 or 1.0e-14 (Report value to the correct number of significant figures.) d) The pH of the solution at the equivalence point is e) On your sheet of paper, draw a reasonable diagram of the complete titration curve. Label both axis appropriately. Label the volumes and pH values at the MIDPOINT and the EQUIVALENCE POINT.