In an experiment, Sadhvi and her lab partner, Nandini, make a buffer containing hydrazoic acid, HN3, and the salt potassium azide, KN3. The concentration of HN3 is 0.3 M, the concentration of KN3 is 0.4 M, and Ka is 2.5 x 10-5. а. Find the pH of the buffer solution if 0.006 mole of HBr are added to 50.0 mL of the buffer. b. Find the pH of the buffer solution if 0.08 mole of LİOH are added to 400.0 mL of the original buffer.

In this lab, we will experimentally determine the titration curve for the titration of a weak acid solution against a strong base solution. The weak acid we will use is potassium hydrogen phthalate, KHP, its chemical structure is shown below. KHP is an ionic compound composed of a potassium cation K+ and a hydrogen phthalate anion HP–. HP– is a weak acid and upon dissolving in water, can lower the pH of the solution. (a)  Suggest the chemical reaction(s) when a solid sample of KHP is dissolved in water, writing out the chemical equations for them. (b)  Sketch the structure of KHP from above and circle the hydrogen atom that is responsible for its acidity. (c)  Calculate the pH of a solution made of 0.50 g of KHP and 50 mL of water. KHP has a molar mass of 204.2 g and at 25 °C has a pKa of 5.4.

A buffer solution is made that is 0.402 M in HF and 0.402 M in KF. If Ka for HF is 7.20 x 10-4 what is the pH of the buffer solution? pH = Write the net ionic equation for the reaction that occurs when 0.107 mol HNO3 is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter. Use H3OT instead of H+) + > +

You have 500.0 mL of a buffer solution containing 0.250 M acetic acid (CH3COOH) and 0.240 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 15.0 mL of 1.00 M NaOH solution? [Ka = 1.80 x 105 for acetic acid.] A) 5.017 B) 4.74 C) 4.657 D) 4.473 Question 25 of 33 E) 4.833

Consider the titration of a weak monoprotic acid with a strong base. The titration starts with 80mL of a weak acid solution with a concentration of 0.160 M. The Ka of the weak acid is 2.9*10^-8. The solution of the weak acid is titrated by adding a 0.290 M NaOH solution. (NaOH is a strong base) A. What is the pH of the solution after 15mL of the base is added?

Phosphoric acid is a triprotic acid (?a1=6.9×10−3 M, ?a2=6.2×10−8 M, and ?a3=4.8×10−13 M). To find the pH of a buffer composed of H2PO4−(aq) and HPO2−4(aq), which pKa value would you use in the Henderson–Hasselbalch equation? 1. p?a1=2.16 2. p?a2=7.21 3. p?a3=12.32   Calculate the pH of a buffer solution obtained by dissolving 21.021.0 g KH2PO4(s) and 39.0 g Na2HPO4(s) in water and then diluting to 1.00 L.   pH=?

A buffer with pH = 8.96 must be created. 8.90 g of the acid component of the best buffer system will be used, and the volume of the solution will be 1 L. Choose which is the best buffer system, then compute the mass of the salt or base component of the chosen system.   Choices:  HClO (Ka = 2.9 x 10-8 , 52.46 g/mol)& KClO (90.55 g/mol)  NH3 (Ka = 5.65 x 10-10 , 17.031g/mol) &NH4Cl (53.491 gmol) CH3COOH (Ka = 1.77 x 10-5 , 60.052 g/mol) & CH3COOK (98.15 g/mol) HClO2 (Ka = 1.1 x 10-2 , 68.46 g/mol)) & KClO2 (106.55 g/mol)

Acetic acid has a Ka of 1.8 x 10¬º. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: A. [acetic acid] ten times greater than [acetate], B. [acetate] ten times greater than [acetic acid], and C. [acetate] = [acetic acid]. Match each buffer to the expected pH. Drag the appropriate items to their respective bins. • View Available Hint(s) Reset Help [acetate] ten times greater than [acetic acid] [acetate] = [acetic acid] [acetic acid] ten times greater than [acetate] pH = 3.74 pH = 4.74 pH = 5.74

a) Calculate the pH of a solution containing 0.30 M HF and 0.30 M NaF.The Ka for HF is 7.2×10-4. pH =  b)  Carbonate buffers are important in regulating the pH of blood at 7.40. If the carbonic acid concentration in a sample of blood is 0.0014 M, determine the bicarbonate ion concentration required to buffer the pH of blood at pH = 7.40. Ka = 4.3×10-7 Сoncentration =  M c) You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH , the pH is 5.50. What is the Ka  value of HA ?  =  d) Consider the titration of 80.0 mL of 0.100 M Ba(OH)2 by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added. 40.0 mL pH =  50.0 mL pH =

A) A buffer solution contains 0.413 M KHSO3 and 0.242 M Na2SO3. Determine the pH change when 0.121 mol KOH is added to 1.00 L of the buffer.pH after addition − pH before addition = pH change B )A 24.0 mL sample of 0.256 M methylamine, CH3NH2, is titrated with 0.335 M hydrobromic acid.After adding 7.50 mL of hydrobromic acid, the pH is . C) A 29.4 mL sample of 0.202 M ethylamine, C2H5NH2, is titrated with 0.295 M nitric acid.At the equivalence point, the pH is . D) A 23.2 mL sample of 0.365 M triethylamine, (C2H5)3N, is titrated with 0.214 M hydroiodic acid.After adding 57.4 mL of hydroiodic acid, the pH is