Calculate the pH of the resulting solution if 34.0mL of 0.340 M HCl (aq) is added to 44.0 mL of 0.390 M NaOH (aq) . Please show steps to solve, formulas utilized and any other applicable changes. I'm not sure what I am doing wrong and want to see how to get to the solution correctly. Will thumbs up! Thanks.

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16.35 A 25.0-mL solution of 0.100 M CH;COOH is ti- of NaOH. trated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL. 16.5 Acid-Base Review Questions

A solution is prepared that is initially 0.17 M in trimethylamine ((CH,), N), a weak base, and 0.20 M in trimethylammonium chioride NHC Complete the reaction table below, so that you could use it to calculate the pH of this solution. Use x to stand for the unknown change in You can leave out the M symbol for molarity. [oH] initial change final Check Explanation 2021 McGraw.HilE olo

You add 7.00 mL of 0.100 M NaOH to 50.00 mL of pure water, and to this mixture you then add 12.00 mL of 0.800 M HCl. What will be the pH of the resulting solution?

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Average HCl 0.421203 Average M NaOH 0.0900235

Calculate the pH of a 0.5 Litre solution wich prepared by adding (50ml NH3) to a sufficient amount of pure water at (25°C). Consider: (at 25°C) Molar mass of NH3: 17g/mol Density of NH3: 0.75kg/m Kw= 1014 K(NH3)= 1.8 x 10s

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.