A 55.0 mL solution of 0.111 M potassium alaninate ( H2NC2HCO2K) is titrated with 0.111 M...
CH3 A 60.0 mL solution of 0.183 M potassium alaninate (H2NC,H,CO,K) is titrated with 0.183 M HCl. The pKa values for the amino acid alanine are 2.344 (pKal) and 9.868 (pKa2), which correspond to the carboxylic acid and amino groups, respectively. H2NCH——0 Calculate the pH at the first equivalence point. Potassium Alaninate pH = Calculate the pH at the second equivalence point. pH = ||
A 50.0 mL solution of 0.174 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.174 M HCl. The pKa values for the amino acid alanine are 2.344 (pKa1) and 9.868 (pKa2), which correspond to the carboxylic acid and amino groups, respectively. a. Calculate the pH at the first equivalence point b. Calculate the pH at the 2nd equivalence point
A 70.0 mL solution of 0.120 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.120M HCL. The pKa values for the amino acid alanine are 2.344 (pKa1) and 9.868 (pKa2), which correspond to the carboxylic acid and amino groups, respectively. a) Calculate the pH at the first equivalence point. pH = b) Calculate the pH at the second equivalence point. pH =
сн. A 75.0 mL solution of 0.146 M potassium alaninate (H, NC,H,CO,K) is titrated with 0.146 M HCl. The pK values for the amino acid alanine are 2.344 (PK) and 9.868 (pK22), which correspond to the carboxylic acid and amino groups, respectively. H2N- CH-C- Calculate the pH at the first equivalence point. Potassium Alaninate pH = Calculate the pH at the second equivalence point. pH =
A 75.0 mL solution of 0.132 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.132 M HCl. The pKa values for the amino acid alanine are 2.344 (pKa1) and 9.868 (pKa2), which correspond to the carboxylic acid and amino groups, respectively. The alaninate ion is the basic form of the diprotic amino acid alanine. The alaninate ion reacts with a strong acid to form the neutral, amphiprotic, intermediate species at the first equivalence point. The intermediate species then reacts with the...
The pKa of hypochlorous acid is 7.530. A 55.0 mL solution of 0.123 M sodium hypochlorite (NaOCl) is titrated with 0.261 M HCI. Calculate the pH of the solution after the addition of 8.24 mL of 0.261 M HCl. pH = Calculate the pH of the solution after the addition of 26.7 mL of 0.261 M HCI. pH = Calculate the pH of the solution at the equivalence point with 0.261 M HCl. pH =
The pK, of hypochlorous acid is 7.530. A 55.0 mL solution of 0.116 M sodium hypochlorite (NaOCI) is titrated with 0.252 M HCI. Calculate the pH of the solution after the addition of 9.37 mL of 0.252 M HCL. pH = Calculate the pH of the solution after the addition of 26.6 ml of 0.252 M HCI. pH = Calculate the pH of the solution at the equivalence point with 0.262 M HCI. pH =
A solution of 0.181 M cysteine is titrated with 0.0453 M HNO_3. The pK_a values for cysteine are 1.70, 8.36, and 10.74, corresponding to the carboxylic acid group, thiol group, and amino group, respectively. Calculate the pH at the equivalence point. pH =
a) A 25.00-mL sample of monoprotic acid was titrated with 0.0800 M potassium hydroxide solution. The equivalence point was reached after 18.75 mL of base was added. Calculate the concentration of the acid. b) A 15.00-mL sample of 0.120 M nitric acid was titrated with 0.0800 M potassium hydroxide. Calculate the pH of the sample when 10.00 mL of the base has been added.
1A solution of 0.235 M aspartic acid, the charge neutral form of the amino acid, is titrated with 0.118 M NaOH . The p K a values for aspartic acid are 1.990 , 3.900 , and 10.002 , corresponding to the α-carboxylic acid group, the β-carboxylic acid group, and the amino group, respectively. Calculate the pH at the first equivalence point of this titration.