a. A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with 0.0919 M NaOH and the equivalence point volume was determined by graphical means to be 25.91 mL. What is the concentration of the acetic acid?
concentration of HA = ___ M
b. Calculate the pKa of a weak monoprotic acid if the pH of a 0.162 M solution is 2.52.
pKa = ___
a. A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with...
A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with 0.1007 M NaOH and the equivalence point volume was determined by graphical means to be 25.42 mL. What is the concentration of the acetic acid? concentration of HA = ___ M
A 0.625-gram sample of an unknown weak acid (call it HA for short) is dissolved in enough water to make 25.0 mL of solution. This weak acid solution is then titrated with 0.100 M NaOH and 45.0 mL of the NaOH solution is required to reach the equivalence point. Using a pH meter, the pH of the solution at the equivalence point is found to be 8.25. Determine the pKa value of the unknown acid.
5 g of an unknown acid, HA, is dissolved in enough water to provide 25.0 mL of solution. The pH of this HA (aq) solution is 2.1 . This solution is titrated with a 0.210 M NaOH solution. 60.2 mL of this NaOH solution is needed to reach the equivalence point. (a) What is the molar mass of HA? (b) What is the value of pKa for HA (aq)? (c) What is the pH at the equivalence point? (d) What...
2.77 g of an unknown acid, HA, is dissolved in enough water to provide 25.0 mL of solution. The pH of this HA (aq) solution is 1.33. This solution is titrated with a 0.250 M NaOH solution. 40.6 mL of this NaOH solution is needed to reach the equivalence point. (a) What is the molar mass of HA? (b) What is the value of pKa for HA (aq)? (c) What is the pH at the equivalence point? (d) What is...
1) A solution of a weak monoprotic acid of unknown concentration was titrated with 0.23 M NaOH. If a 100.-mL sample of the acid solution required exactly 10. mL of the NaOH solution to reach the equivalence point, what was the original concentration of the weak acid? 2) During the titration on problem (2B), after 5.0 mL of NaOH addition, the pH = 3.68. What is the Ka of the weak acid? please show steps i have an exam tomorrow
2.77 g of an unknown acid, HA, is dissolved in enough water to provide 25.0 mL of solution. The pH of this HA (aq) solution is 1.33. This solution is titrated with a 0.250 M NaOH solution. 44.2 mL of this NaOH solution is needed to reach the equivalence point. (a) (2 marks) What is the molar mass of HA? (b) (2 marks) What is the value of pKa for HA (aq)? (c) (2 marks) What is the pH at...
A 25.0 mL sample of an acetic acid solution (Ka = 1.76 × 10-5) is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid is __________ M.
Calculate the pH of a 25.0 mL of 0.100M base acetic acid solution after being titrated with 0.100 M NaOH to its equivalence point (pKb (acetic acid)=5.68x10^-10)
A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The titration required 30.0 mL of base to reach the equivalence point, at which point the pH was 8.68. a) What is the molecular weight of the acid? b) What is the pKa of the acid?
A 25.0 mL NaOH solution of unknown concentration was titrated with a 0.189 M HCl solution. 19.6 mL HCl was required to reach equivalence point. In a separate titration, a 10.0 mL H3PO4 solution was titrated with the same NaOH solution. This time, 34.9 mL NaOH was required to reach the equivalence point. What is the concentration of the H3PO4 solution?