CH3 A 60.0 mL solution of 0.183 M potassium alaninate (H2NC,H,CO,K) is titrated with 0.183 M...
A 55.0 mL solution of 0.111 M potassium alaninate ( H2NC2HCO2K) is titrated with 0.111 M HCI. 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. Calculate the pH at the first equivalence point. CH3 H2N Potassium Alaninate 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 a hypochlorous acid is 7.530. A 60.0 mL solution of 0.138 M sodium hypochlorite (NaOCl) is titrate with 0.319 M HCl. Calculate the pH of the solution: a) after the addition of 7.79 mL of 0.319 M HCl b) after the addition of 27.2 mL of 0.319 M HCl c) at the equivalence point with 0.319 M HCl
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.
Consider the titration of 50.0 mL of 0.183-M of KX with 0.090-M HCl. The pKa of HX = 8.13. Give all pH values to 0.01 pH units. a) What is the pH of the original solution before addition of any acid? pH = 10.70 b) How many mL of acid are required to reach the equivalence point? VA = 101.67 mL c) What is the pH at the equivalence point? pH = 4.67 d) What is the pH of the...
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 0.1983 g sample of Na2CO3 was dissolved in 100.00 ml H20. That solution was titrated with 0.1531 M HCL as titrant. The PKa1 and PKa2 of carbonic acid (the conjugate acid of the carbonate ion) are 6.351 and 10.329, respectively. 1.) What is the initial pH of the sodium carbonate solution. 2.) What is the first equivalence point (in ml HCL)? What is the pH at 1st eq.? 3.) What is the second equivalence point (in ml HCL)? What...