lab

? ? = 𝑉 ? 𝐶 ? = 𝑉 ? 𝐶 ? = ? ? Trial 1: Volume NaOH = 38.2 mL = 0.038 L Moles of KHP = 0.00376 mol 𝐶 ? = ? ? 𝑉 ? = 0.00376??? 0.038 ? = 0.0985 ??? ? Trial 2: Volume NaOH= 42.5 mL = 0.0425 L Moles of KHP = 0.004181 mol 𝐶 ? = ? ? 𝑉 ? = 0.004181??? 0.0425 ? = 0.0984 ??? ? Trial 2: Volume NaOH= 42.5 mL = 0.0425 L Moles of KHP = 0.004181 mol 𝐶 ? = ? ? 𝑉 ? = 0.004181??? 0.0425 ? = 0.0984 ??? ? The concentration of NaOH is 0.09835 M Questions for Part II 2. Calculate the molecular weight of the diprotic acid in g/mol using the mass of diprotic acid that you measured in the first step of the procedure and the moles you determined from the titration results (Show all your work). Finding moles of acid (first equivalence): ? ? = 𝐶 ? 𝑉 ? = (0.09835?) ( (0.0367 ?) 2 ) = 0.0018047 ??? Finding molecular: weight: ?? = ? ? ? ? = 0.124𝑔 0.001844??? = 68.71 𝑔 ??? 3. From the following list of five diprotic acids, identify the unknown diprotic acid that you used. Note that not everyone uses the same acid. Diprotic Acid Formula MW Oxalic Acid H 2 C 2 O 4 ·H 2 O 126 Malonic Acid H 2 C 3 H 2 O 4 104 Maleic Acid H 2 C 4 H 2 O 4 116 Malic Acid H 2 C 4 H 4 O 5 134 Tartaric Acid H 2 C 4 H 4 O 6 150 The unknown acid could potentially be Malonic Acid.