In an analysis of metal carbonate M2CO3∙3H2O, 1.500g of the unknown carbonate was dissolved in enough water to make a 100.0mL solution. 20.00mL of such solution was treated with 10.00mL of 1.000M HCl solution, and the excess acid was titrated with 54.92mL of 0.1138M NaOH solution. The equations are given below:     CO32-(aq) + 2 H+(aq) → H2O(l) + CO2(g)     H+(aq) + OH-(aq) → H2O(l)   a) the moles of NaOH used in the titration is (enter the answer in 4 sig. figs.) Answer       mol   b) the mole of HCl reacted with the carbonate is (enter the answer in 4 sig. figs.) Answer       mol   c) the molar mass of M2CO3∙3H2O is (enter the answer in 4 sig. figs.) Answer     g/mol   d) the identity of metal M is (enter the element symbol) Answer

You are performing an experiment in lab that involves the titration of 25.0 mL of H2SO4 solution. You titrate the acidic solution with 0.8221 M NaOH and the equivalence point is reached by the addition of 16.25 mL of NaOH solution. Using the balanced equation below, calculate the Molarity of H2SO4 in the flask. Do NOT include units. 2NaOH(aq) + H2SO4(aq) → 2H2O(l) + Na2SO4(aq)

A chemist needs to determine the concentration of a solution of nitric acid, HNO3. She puts 905 mL of the acid in a flask along with a few drops of indicator. She then slowly adds 0.200 mol L Ba(OH)2 to the flask until the solution turns pink, indicating the equivalence point of the titration. She notes that 225 mL of Ba(OH)2 was needed to reach the equivalence point. Solution map In this titration, the concentration of base is known and can be used to calculate the unknown acid concentration: concentration of base → moles of base → moles of acid → concentration of acid Part A How many moles of Ba(OH)2 are present in 225 mL of 0.200 mol L-1 Ba(OH)2? Express your answer numerically in moles. • View Available Hint(s) Vo AEO ? mol Ba(OH)2 Submit

A solution of H2C2O4 is used to titrate a 18.42 mL sample of a 0.2079 M KOH solution. If the titration requires 13.51 mL of the H2C2O4 solution, what is its molarity?    I need the work to understand it

A strong acid-strong base titration was performed as follows: 10.00 mL of 0.10M HCl were collected in an Erlenmeyer flask, and then diluted to 100 mL with distilled water. After that, two drops of indicator were added. Then, the buret was filled with 0.100 M NaOH. The titration was started by slowly adding NaOH to HCl. (Note: The volume of two drops indicator is ignored as it is so tiny compared with 100 mL.) What is the pH of the solution in the Erlenmeyer flask after adding 10.20 ml of 0.100M NaOH? Select one:a. 9.80b. 10.26c. 9.90d. 11.08

A student prepares a dilute solution of sodium hydroxide, NaOH (aq), starting with 6 M sodium hydroxide. She then titrates a 1.372 g sample of KHP with the dilute sodium hydroxide solution, NaOH (aq), to a phenolphthalein end point.   A.) If the titration required 21.84 mL of sodium hydroxide, NaOH (aq), calculate the molar concentration of the sodium hydroxide solution, NaOH (aq). (Remember that KHP is potassium hydrogen phthalate, KHC8H4O4, NOT potassium hydrogen phosphorus!)   B.) The student uses the same sodium hydroxide to titrate 10.00 mL of vinegar to a phenolphthalein end point. If the titration required 27.48 mL of sodium hydroxide, NaOH (aq), calculate the molar concentration of acetic acid, HC2H3O2 (aq), in the vinegar.   C.) Calculate the mass percent of acetic acid, HC2H3O2 (aq), in the vinegar using the molar concentration for acetic acid, HC2H3O2 (aq), determined in part b and assuming the density of the solution is 1.01 g/mL.

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation:HA(aq)+KOH(aq) --------->KA(aq)+H2O(l)       If 14.90 mL of 0.585 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? ______ M      What is the molar mass of HA? _____  g/mol

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50 mL of a 1.067 M solution of HCI was diluted to 250 mL. The diluted HCI solution was then titrated with an unknown concentration of KOH. It is found that an average volume of 24.20 mL of KOH is required to reach an endpoint in a titration with 25.00 mL of the dilute acid solution. The equation for the reaction is; H(aq) + OH(aq) →→> H₂O (1) The concentration of KOH is 0.2205 M Calculate the concentration of the KOH in g/L (you will need to find the formula weight of KOH to do this) 15| ΑΣΦ ? g/L

An analytical chemist weighs out 0.318 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.0600 M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 82.8 mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits. mol x10 x 3