If 0.259 mol of a monoprotic weak acid (pKa = 4.03) are dissolved in water and titrated with NaOH, what will the pH be at the half-equivalence point? Report your answer to the hundredths place.
Let the monoprotic acid be HA.
At half equivalence point [HA] = [A-]
Therefore, ka = [H3O+]
Hence, at half equivalence point pH = pKa of acid
So, pH at half equivalence point will be 4.03.
If 0.259 mol of a monoprotic weak acid (pKa = 4.03) are dissolved in water and...
The pka of a weak acid is useful for identification of an unknown acid because it is a constant (as long as temperature remains the same). When a weak acid is titrated with a strong base, the pka of the weak acid equals the pH at the half-equivalence point. A 0.3210 g sample of weak monoprotic acid is delivered into laboratory glassware. It will be used for titratation against 0.1254 M NaOH. If the molar mass of the weak acid...
A sample of 0.2140 grams of an unknown monoprotic weak acid was dissolved in 25.0mL of water and titrated with 0.0950M NaOH. The acid required 15.50mL of NaOH to reach the equivalence point. What is the molar mass of the unknown acid?
3) A weak monoprotic acid has a pKa 6.15. 50.00 mL of an 0.1250M aqueous solution of this weak acid is titrated with 0.1000M NaOH. a) What is the equivalence point volume and 2 equivalence point volume for this titration? Find the pH b) before the titration begins; c) after 20.00 mL of the NAOH has been added, d) after 62.50 mL of the NaOH has been added; and e) after 85.00 mL of the NAOH has been added. 4)...
A sample of 0.2140 g of an unknown monoprotic weak acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 15.50 mL of NaOH to reach the equivalence point. What is the molar mass of the unknown acid?
2. A 0.500-gram sample of a weak, nonvolatile monoprotic acid, HA, was dissolved in sufficient water to make 50.0 milliliters of solution. The solution was then titrated with a standard NaOH solution. Predict how the calculated molar mass of HA would be affected (too high, too low, or not affected) by the following laboratory procedures. Explain each of your answers. A. (2 points.). After rinsing the buret with distilled water, the buret is filled with the standard NaOH solution; the...
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 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.
If 0.360 moles of a monoprotic weak acid (Ka = 8.6 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.773 Volume of NaOH required to reach endpoint: (ml) 19.0 pH of the mixture Ihalf neutralized solution 3.54 Calculate the following...
12.22 g of the monoprotic acid KHP(MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.85 M solution of calcium hydroxide to the equivalence point. What volume of base was used?