A plot of pH vs volume of 0.05 M NaOH solution was obtained for titrating 0.4055 g of a dried solid sample of a weak, monoprotic acid.    Determine the molar mass of the acid in its solid, crystalline form.

The student Louise continued the analysis of acetic acid using half - neutralization technique. She titrates this acetic acid solution using 0.984 M NaOH titrant solution. When the volume of NaOH reaches 14.34 mL, she stopped titration. The averaged pH value is measured to be 4.75.  - Calculate the molar concentration of H3O ^+.

A solution of the weak acid HA is prepared by dissolving 2.70 g of HA in 100.0 mL water. The solution is titrated, and the equivalence point is reached after 31.0 mL of 0.500 M NaOH is dispensed. Calculate the molar mass of HA. molar mass: g/mol At the equivalence point in this titration, how can the solution pH be described? acidic neutral basic

a. A student massed out 0.175 g of an unknown diprotic acid on an analytical balance, and then dissolved it in 100mL of deionized water. The acid was then titrated with 0.111 M NaOH solution. The second equivalence point showed the sharpest change in pH, and so it was used to determine the molar mass of the unknown acid. The volume of NaOH needed to reach the second equivalence point was 27.77 mL   b. If the volume required to reach the second equivalence point is 38.18 mL, what is the volume required to reach the first equivalence point?   Report your answer to 2 digits behind the decimal point, and do not include units in your answer.   c. if the volume to reach the second equivalence point in a titration is 24.53 mL, What is the volume to reach the midpoint between the first equivalence point and the second equivalence point?   Report your answer to 2 digits behind the decimal point and do not include units   d. If pKa1 = 3.18 what is the value for Ka1?   Report your answer in…