KHP(aq) + NaOH(aq) -> H2O (l) + KNaP(aq) 1. Once the equivalence point is reached, any NaOH that is added after that reacts with the indicator giving the pink color. If too much NaOH is added the pink color is too dark. How would this affect the value of the calculated NaOH molarity?   2. Why might it be a good idea to rinse down the sides of the flask with a squirt of deionized water from time to time during the titration?   3. When a little NaOH is added to the KPH solution, why is a pink color initially seen, but then it disappears?

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…

Does it matter how much water is used to dissolve the unknown monoprotic acid? Explain your answer.     Yes.  Adding a higher (or lower) amount of water would change the number of moles of the unknown monoprotic acid in the solution, so the equivalence point will be reached at different pH.         Yes.  Adding a higher (or lower) amount of water would change the initial concentration of unknown monoprotic acid, so the equivalence point will be reached at different pH.     No.  Adding water does not change the number of moles of the unknown monoprotic acid, so the amount of water is insignificant.

. A 4.59 mL sample of HCl, specific gravity 1.3, required 50.5 mL of 0.9544N NaOH in a titration. Calculate the % w/w HCl.

A sample of acetic acid is titrated with a standardized NAOH solution. Before the end-point of the titration, which of the following must be true? HC,H3O, (aq) + NAOH (aq) → NaC,H3O2 (aq) + H2O (1) Note: [X] = molar concenctration of X %3D O [H+] > [OH-] [-HO] > [+H] O %3D [-HO] = [+H] O

Consider the titration of 40.0 mL of 0.200 mol/L HCOONa (aq) with 0.300 mol/L HCI (aq). Ka for HCOOH is 1.8×104. Determine the pH of the original solution. Use the given values to set up an ICE table. Determine the unknown equilibrium concentrations. Initial (M) Change (M) Equilibrium (M) 1 HCOONa(aq) H₂O(1) HCOOH(aq) OH (aq) 5 RESET

Calculate the amount of 6M HCl required to neutralize the solutions in beakers #1 and #2. Note: Saturated NaHCO3 has a concentration of about 1.1M. (use the last page for your calculations)

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Determine the molar concentrations of Ca2+ and CI- in a 0.38 M calcium chloride solution. 1) 0.38 M Ca2+ and 0.38 M CI- O 2) 00.19 M Ca2+ and 0.38 M CI- 3) 0.38 M Ca2+ and 0.76 M CI- 4) 0.19 M Ca2+ and 0.76 M CI- O 5) None of the above

2. In the second titration, the molar mass of an unknown monoprotic acid is calculated.  If 0.100 grams of unknown solid are dissolved in water and titrated with 4.68 mL of the NaOH, what is the molar mass of the solid? HINTS: Molar mass is grams / moles. We have grams from the weighed mass, we can calculate moles of acid from the titration. Monoprotic means that it produces 1 H+ per molecule. This means that our moles of NaOH added and moles of acid are the same. You will need to use your concentration of NaOH from the previous question Which is 0.1186