You are given a bottle of NaOH of an unknown concentration. You are tasked with determining the concentration of this solution. You weigh out 0.3175 g of KHP (MW = 204.2 g/mol) and dissolve it in 40 mL of water. You add phenolphthalein to the solution as an indicator. You preform your titration and determine that it requires 19.45 mL of the unknown NaOH solution to reach the endpoint. From this information, calculate the concentration of the unknown solution.​

0.07975 M NaOH

Explanation:

First determine the molarity of your acid, KHP. Convert grams to moles using the MW given and then divide by the volume of water.

0.3175 g KHP • (1 mol KHP / 204.2 g KHP) = 0.001555 mol KHP

0.001555 mol KHP / 0.040 L = 0.03888 M KHP

The reaction between KHP and NaOH is one-to-one, so you can just use M1V1 = M2V2 to solve for M2 (the molarity of NaOH).

M1 = 0.03888 M KHP

V1 = 40 mL

M2 = ? M NaOH

V2 = 19.45 mL NaOH

(0.03888 M KHP)(40 mL) = M2(19.5 mL NaOH)

M2 = 0.07975 M NaOH

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g