Titrations are acid-base reactions in which the amount of acid and base in solution are known quite precisely. The specific case wherethe acid and base have been added in equivalent molar amounts, is called the equivalence point. At this point,mol acid = mol base%3DAt all other points in a titration, either the base is the limiting reactant (and there is excess acid) or the acid is the limiting reactant (andthere is excess base). So, except at the equivalence point, titration problems are limiting reactant problems, which means we can dothem- a lot of them!One additional point: when there is left over strong acid (acid in excess), 100% of that excess acid is ionized to become H* ions:[strong acid] = [H*] and pH = -log[H"]Something similar happens for strong bases.1. Calculate the pH of the solution after the following amounts of 0.250 M NaOH have been added to 50.0 mL of 0.200 M HCI. Showall your work for each calculation (including the volumes for which the pH is given). Feel free to attach additional pages to yoursubmission if needed.mL of NaOHpH0.00ANS: 0.69936.0039.7540.0040.2544.0066.00ANS: 12.752. The point at 40.00 mL of NaOH is a turning point in the calculations. How so? What should the pH be at/40.00 mL and why? Howabout after 40.00 mL?

HCl and NaOH react by neutralisation reaction HCl +NaOH→NaCl +H2O. Moles of HCl present =Molarity…

Answered: 1. Calculate the pH of the solution…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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pH calculation after first equivalence point: Calculate the pH of 50.0 mL of 0.250 M oxalic acid (HOOCCOOH) upon addition of 48.6 mL of 0.500 M NAOH. Round your calculated value for pH to two figures to the right of the decimal point. Significant constants: Ka, of oxalic acid = 5.60 x 10 and Ka, of oxalic acid = 5.42 x 10
What volume of 0.837 M KOH solution do you need to make 3.62 L of a solution with a pH of 12.700? Express your answer to three significant figures and include the appropriate units. HÅ V-21.6 4 ml ENC Submit Previous Answers Bequat Answer ? X Incorrect; Try Again; 4 attempts remaining Begin by calculating [OH-] in the final solution using the given pH value. Since KOH is a strong base and completely dissociates into K and OH in an aqueous solution, the molarity of OH equals the molarity of KOH. Finally, rearrange the dilution equation (M₁ V₁M₂V₂) to solve for the initial volume (Vi).
18. Calculate the pH of a solution created by mixing 50.00 mL of 0.200 M HF with 25.00 mL of 0.250 M NaF.
Calculate the pH of the solution after 433 mL of 0.561 M HCl is added to 2.43 L of a solution that is 0.226 M HC2H3O2 and 0.416 M KC2H3O2. Ka for HC2H3O2 is 1.8x10^-5
4. What volume, in mL of 0.197 M HCIO4 needs to be added to 187.2 mL of 0.546 M RBCIO to achieve a solution with a pH of 7.35. Note: the pkb of RBCIO is 6.47, you may assume the 5% approximation holds, and you may assume that the volumes are additive. INSTRUCTIONS: Input your answer to 4 significant digits in standard notation (example: 1.234) and DO NOT include the units. mL Submit Answer Tries 0/99
... 2@ 48 TO Time left 0:24:03 Consider a solution initially containing 0.40 mol fluoride anion (F) and 0.30 mol of hydrogen fluoride (HF). If 0.40 mol of HCl are added to this solution, which of the following statements is/are FALSE? Addition of 0.40 mol of HCI will lover the pH of solution by 0.4 unit. O You'll essentially have a weak acid solution situation, with 0.7 mol HF at the end. O You will still have a buffer solution at the end, since you'll still have significant amounts of both weak base and conjugate weak acid. You will no longer have a buffer solution, since all of the buffer capacity was exhausted. The pH will have shifted to a lower pH.
Complete the data sheet below. DETERMINATION OF pH OF DIFFERENT SOLUTIONS For those who will be using scientific notation, please follow the format below: Example: 0.00455 -> 4.55E-3 45000 -> 4.5E+4 SOLUTION pH [H*), M pOH [он], м 0.0345 M HCI 0.000890 M Sr(OH)2 Mixture of 10.0 mL each of the 2 solutions above
A chemist titrates 150.0 mL of a 0.6575M hydrobromic acid (HBr) Solution with 0.4987M 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 =
The pH of a 1.00 M NaCH3COO (aq) solution is 9.38. a/ Determine the concentration of [OH-] in the solution. b/ Write the expression for the Kb of NaCH3COO (aq) and calculate its values. c/ Calculate the pKa of CH3COOH. d/ HCl (aq) is added to 1.00 L solution of 1.00 M NaCH3COO (aq). Write the chemical reaction between HCl (aq) and NaCH3COO (aq). e/ How many mol of HCl (aq) is needed to make a buffer solution with pH = 5.00 from 1.00 L of 1.00 M NaCH3COO (aq).
Please answer these questions on your Page 9 We dissolve 2.66 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution is 1.11. We titrate this solution with a 0.166 M NaOH solution and we need 22.2 mL of this NaOH solution to reach the equivalence point. The temperature is 25.0°C throughout. (a) What is the molar mass of HA? (b) What is the dissociation constant, K₁, of A-(aq)? (c) What was the pH of the solution after the addition of the first 15.0 mL of the NaOH solution?
This question has two parts. Part-I (a) Calculate the pH of a 0.20 M HF solution. The Ka of HF is 6.8*10-4. Show the ICE table and all calculations. Part-II (b) Calculate the pH of a solution containing both 0.20 M HF and 0.10 M HCl? The Ka of HF is 6.8*10-4. Show the ICE table and all calculations.
calculate the ph of each of the following volumes of .180 M KOH added to 20.0 mL of .216 M HA (Ka = 4.3x10^-5).a) 0 mLb) 9 mLc)12 mLd) 18 mLe) 24 mLf) 32 mL

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