EXPERIMENT 5; Acid-Base Titration A. Trial 1 Trial 2 Trial 3 Mass of KHP 5.210 g 5.210 g 5.210 g (C3H;O4) Mole of KHP ---------- Initial Burette 0.00 0.50 5.00 Volume NaOH (mL) Final Burette 31.00 26.25 29.95 Volume NaOH (mL) Total Volume of NaOH (mL) Moles of NaOH Molarity of NaOH (mol/L) Calculate the average Molarity of NAOH

please help me with both. I keep getting it wrong

Question Completion Status: For Solution A, approximately how many mL of HCI were needed to exceed the buffering capacity of this solution? Enter the number only. Don't enter units as part of your answer and don't use commas or scientific notation. Titration curve of Solution A 14 13 12 11- 10 9 8 pH 7 6 5 4 3 2 1 5 4 3 2 1 HCI added (mL) 0 1 2 3 4 5 NaOH added (mL) 14 13 12- 11 10 9 8 pH 7 6.8 2 65 4 3 2 1 Titration curve of Solution B Click Save and Submit to save and submit. Click Save All Answers to save all answers. 54 32 1 HCI added (mL) 0123 NaOH added (mL)

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 HCI have been added. a. 0.0 mL pH = b. 20.0 mL pH = c. 30.0 mL pH= d. 40.0 mL PH= e. 90.0 mL pH = Submit Answer Try Another Version 1 item attempt remaining

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Part C. Determination of Unknown Balanced chemical equation: Hcl+ Na OH -o Nacl t Hg0 Unknown number NA Trial 1 Trial 2 Trial 3 Volume of unknown HCl titrated 10.00mL NA NA Initial burette reading 0.90mL Final burette reading 30.90mL Volume NaOH added 20 00ML 50.00MLXO.144 m=43.2m moles NaOH added 4.335m (Use average molarity from part B and multiply it by the volume you added) 0.4335 M moles HCl 4.335m 04335m 10.00 mL molarity of HCl Actual Molarity (from instructor) 0.3150M percentage error

2. A titration was performed by adding 0.1230 M HNO3 to 25.0 mL of 0.01673 M KOH. What is the pH of the solution at the equivalent point? a. pH >7; b. pH = 7; c. pH <7 d. cannot determine

How many milliliters of water have to be added to 143.8 mL of 0.25 M HCI to reduce the concentration to 0.10 M HCI? Vwater = mL eTextbook and Media Hint Save for Later Attempts: 0 of 2 used Submit Ans O Type here to search 10 insert @ %23 $ & * 4 7 8. 10 %3D Q W R PI A G K C V B NM pause alt alt ctri

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What is the pH at the equivalence point in the titration of 60.00 mL of 0.200 M HA (a generic acid with K, = (2.26x10^-5)) with 0.400 M NaOH? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer