A 91.0 mL sample of 0.0300M HBr is titrated with 0.0600M KOH solution, Calculate the pH after the following volumes of base have been added. (a) 14.6 ml. (b) 45.0 ml (c) 45.5 ml. pH (d) 46.9 ml. (e) 75.1 ml pH
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- Chemistry A 94.0 mL sample of 0.0300 M HIO, is titrated with 0.0600 M NaOH solution. Calculate the pH after the following volumes of base have been added. (a) 14.1 mL pH = (d) 47.9 ml pH- (b) 46.1 mL pH = (e) 90.7 ml pH = (c) 47.0 mL pH =Calculate the pH change that results when 15 mL of 2.7 M HCI is added to 600. mL of each of the following solutions. Use the Acid-Base Table. (a) pure water 4.0 -5.82 (b) 0.10 M CH3COO- 4.0 4.43 (c) 0.10 M CH3COOH 4.0 X (d) a solution that is 0.10 M in each CH3COO and CH3COOH. 4.0A 71.0 mL sample of 0.0100 M HClO4 is titrated with 0.0200 M NaOH solution. Calculate the pH after the following volumes of base have been added. (a) 13.8 mLpH = (b) 35.1 mLpH = (c) 35.5 mLpH = (d) 36.6 mLpH = (e) 61.4 mL
- A 92.0 mL sample of 0.0400 M HBr is titrated with 0.0800 M NaOH solution. Calculate the pH after the following volumes of base have been added. (a) 23.0 mLpH = (b) 45.5 mLpH = (c) 46.0 mLpH = (d) 47.8 mLpH = (e) 81.9 mLA 93.0 mL sample of 0.0100 M HIO4 is titrated with 0.0200 M LIOH solution. Calculate the pH after the following volumes of base have been added. (a) 14.4 mL pH = 2.57 (d) 47.4 mL pH = X X (b) 44.2 mL pH = (e) 78.1 mL pH = X X (c) 46.5 mL pH = XCalculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution after the following additions of base: (a) 28.00 mL (b) 39.50 mL (c) 52.00 mL
- (a) A 50.0 mL solution is prepared to be 1.29 M acetylsalicylic acid. In the first step, 8.55 mL of NaOH is titrated into the solution until the pH is exactly 5.0. What is the concentration of the titrant (NaOH)? (b) In the second step, enough 5.85 M nitric acid is added to the solution after the titration in part (a) is complete until the pH is one unit lower than the pKa of acetylsalicylic acid. What volume (mL) of nitric acid was added?Calculate the pH change that results when 15 mL of 2.0 M HCI is added to 580. mL of each of the following solutions (a) pure water 4.0-5.70 (b) 0.10 M CH3COO 4.04.28 (c) 0.10 M CH3COOH 4.0 (d) a solution that is 0.10 M in each CH3COO and CH3COOH. 4.0A 81.0 mL sample of 0.0500 M HBrO4 is titrated with 0.100 M NaOH solution. Calculate the pH after the following volumes of base have been added. (a) 14.2 mLpH = (b) 39.3 mLpH = (c) 40.5 mLpH = (d) 42.5 mLpH = (e) 79.0 mLpH =
- Calculate the pH change that results when 11 mL of 5.1 M NaOH is added to 790. mL of each the following solutions. Use the Acid-Base Table. (a) pure water 4.0 5.84 (b) 0.10 M NH4CI 4.0 5.43 (c) 0.10 M NH3 4.0✔ X (d) a solution that is 0.10 M in each NH4+ and NH3 4.0✔Calculate the pH change that results when 14 mL of 2.0 M HCI is added to 550. mL of each of the following solutions. (See the Acid-Base Table.) (a) pure water 1.304 X (b) 0.10 M CH3COO- (c) 0.10 M CH3COOH (d) a solution that is 0.10 M in each CH3COO and CH3COOH.A 76.0 mL sample of 0.0100 M HIO4 is titrated with 0.0200 M RbOH solution. Calculate the pH after the following volumes of base have been added. (a) 14.1 mL pH = 3.32 (d) 39.9 mL pH = 9.58 X X (b) 37.2 mL pH = 4.79 (e) 69.5 mL pH = 10.79 X X (c) 38.0 mL pH = 7