hello, can you please help me with this, and explained it by step. Thank you<3 Exercise 4: What weight of sodium formate (NaHC00, 68.008g/mol) must be added to 400.0 ml of 1.00 M formic acid(HCOOH) to produce a buffer solution that has a pH of 3.50? Kafor formic acid = 1.80 x 104.CNAApH = pka + logСНА

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Exercise 4: What weight of sodium formate (NaHCo0, 68.008 g/mol) must be added to 400.0 ml of 1.00 M formic acid (HCOOH) to produce a buffer solution that has a pH of 3.50? K, for formic acid = 1.80 x 104. CNAA pH = pka + log СНА

Exercise 4: What weight of sodium formate (NaHCo0, 68.008 g/mol) must be added to 400.0 ml of 1.00 M formic acid (HCOOH) to produce a buffer solution that has a pH of 3.50? K, for formic acid = 1.80 x 104. CNAA pH = pka + log СНА

General Chemistry - Standalone book (MindTap Course List)

11th Edition

ISBN:9781305580343

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Chapter16: Acid-base Equilibria

Section16.1: Acid-ionization Equilibria

Problem 16.2E: What are the concentrations of hydrogen ion and acetate ion in a solution of 0.10 M acetic acid,...

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hello, can you please help me with this, and explained it by step. Thank you<3

Exercise 4: What weight of sodium formate (NaHC00, 68.008
g/mol) must be added to 400.0 ml of 1.00 M formic acid
(HCOOH) to produce a buffer solution that has a pH of 3.50? Ka
for formic acid = 1.80 x 104.
CNAA
pH = pka + log
СНА

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Calculate the pH of the solution prepared by dissolving 0.48 mol of NH3(Kb=1.77x10-5) and 0.750 mol of ammonium chloride in enough water to make 1.00L of buffer solution 9.44 10.1 no correct answer 9.05 O 5.09
A 5.00 L solution is 0.100M in benzoic acid (Ka=6.3 x 10_5 ) and 0.200M in calcium benzoate. What is the pH of this solution after 5 ml 10.0M HCl are added? Answer: 4.75
Incorrect Question 5 Which of the following mixtures would result in a buffered solution? Mixing 100.0 mL of 0.100 MHC1 with 100.0 mL of 0.100 M NaOH. Ⓒ Mixing 100.0 mL of 0.100 MNH3 (K=1.8 x 105) with 100.0 mL of 0.100 M NaOH. Mixing 100.0 mL of 0.100 MHCI with 100.0 mL of 0.100 MNH3; (K=1.8×10) Mixing 50.0 mL of 0.100 MHCI with 100.0 mL of 0.100 MNH3 (K₂=1.8 × 10-). O At least two of the above mixtures would result in a buffered solution.
Practice Exercise Which of the following couples (b) KBr/HBr, (c) Na,CO3/NaHCO3. are buffer systems? (a) KF/HF,
Exercise: Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 1×10² Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H"]=[HCI] (b) After initial addition of NaOH and before the end point: [H]=MVMNVN V/ LV (c) At the end point (Vsaou 50 mL), all the HC1 is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1-10 (d) After the end point: 1x 10-¹ XV VO+VIRY [H"]= NOW ABRIREDIC titrant: NaOH, 0.1M Question 1) What is the pH when the solution volume in the flask reaches 95 ml? a. 12.79 b.12.82 c. 1.95 d. 12. 75e. 2.28 Question2) What is the maximum pH reached at the end of titration? a. 12.77b. 12.79 c. 12.81 d. 12. 78 e. 12.82 analyte: HC1, 0.1 M, 50 ml.
Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H*] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 1x10¹ Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H] = [HCI] (b) After initial addition of NaOH and before the end point: [H]=MCVMNam *V N₂ V LV (c) At the end point (Vsou 50 mL), all the HCI is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1×10 (d) After the end point: [H"]= 1x10-14 xV VO+VY M NOW ɔ CLOTHING titrant: NaOH, 0.1M Question 1) What is the pH when the solution volume in the flask reaches 95 ml? a. 12.79 b. 12.82 c. 1.95 d. 12. 75e. 2.28 Question2) What is the maximum pH reached at the end of titration? a. 12.77b. 12.79 c. 12.81 d. 12.78 e. 12.82 analyte: HC1, 0.1 M, 50 ml.
Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 $2.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 What is the pH when the solution volume in the flask reaches 95 ml? O a 1279 Ob 12.82 OC 1.95 O d. 12.75 De 2.28 Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H"]=[HCI] (b) After initial addition of NaOH and before the end point: MC VC-M NOVNO [H"]= V LV (c) At the end point (Vou 50 mL), all the HCI is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1-10 (d) After the end point: 1x10-¹4 M Isel Boy XV V₁W+VIRY [H"]= titrant: NaOH, 0.1M analyte: HC1, 0.1 M, 50 ml.
How would you prepare 1.40 L of a buffer with a pH of 6.90 from 0.360 M H3PO4 and 0.300 M NaOH? Mix mL of 0.360 М НЗРОД with mL of 0.300M NaOH. Submit Answer Try Another Version 6 item attempts remaining
Practice Exercise Calculate the pH at the equivalence point in the titration of 50 mL of 0.10 M methylamine (see Table 15.4) with a 0.20 M HCl solution.
Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H'] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 1×107 Hints: The pH of the solution is calculated using the following equations: [H]= (a) Initial conditions (VNaOH = 0), [H] = [HCI] (b) After initial addition of NaOH and before the end point: [H]= MHCI VHCI-M VHCI + V NaOH 1x10-14 XV. titrant: NaOH, 0.1M (c) At the end point (VNaOH = 50 mL), all the HCI is being consumed and the acidity of the solution comes from auto ionization of water which is: [H"]=1x107 (d) After the end point: Mo analyte: HC1, 0.1 M, 50 ml. NaOH *V NaOH NaOff V +V₂C² NaOH
You need to prepare an acetate buffer of pH 5.19 from a 0.747 M acetic acid solution and a 2.49 M KOH solution. If you have 880 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.19? The p?a of acetic acid is 4.76. Be sure to use appropriate significant figures. Please type answer note write by hend.
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