de-ionized water to a final volume of 100.0 mL (10) A 500 mL sample ofabattery f
ID: 1034500 • Letter: D
Question
de-ionized water to a final volume of 100.0 mL (10) A 500 mL sample ofabattery fluid is diluted Then a 10.0-mL portion of the diluted battery solution is pipetted into an Erlenmeyer flask and titrated with NaOH solution of known concentration using phenolphthalein indicator. The acid-base reaction occurs as follows: B5 Na,so,(aq) 2H2O(aq) + H,SOdaq) + 2NaOH(aq) ? (a) If 32.6 mL of 0.128 M NaOH solution were required to reach end-point, how many moles of NaOH were consumed in this titration? (b) How many moles of H,SO, were present in the dilute acid sample titrated? What is the mol concentration of sulfuric acid in diluted battery fluid solution? 13.a .8530 olay nuntrahn (e) Calculate the molar concentration of H,SO, in the original (undiluted) battery fluid. ahaExplanation / Answer
(a) Given
Volume of NaOH = 32.6 ml = 0.0326 L
Molarity of NaOH = 0.128 M ( mol/L)
No. of moles of NaOH = Volume * molarity = 0.0326 L * 0.128 mol/L = 4.173 * 10-3 moles Answer (a)
(b) according to reaction stoichiometry
No. of moles of H2SO4 = 2 * No. of moles of NaOH = 2 * 4.173 * 10-3 moles = 8.35 * 10-3 moles Answer
sample that was titrated = 10 ml = 0.01 L = Volume
Molarity of dilute H2SO4 sample titrated = No. of moles / volume = 8.35 * 10-3 moles / 0.01 L = 0.835 mol/L or M
Answer (b) 8.35 * 10-3 moles of H2SO4 and 0.835 M H2SO4
(c) Given
orignal solution volume V1 = 5 ml = 0.005 L
diluted solution volume V2 = 100 ml = 0.1 L
diluted solution molarity M2 = 0.835 mol/L
V1 * M1 = V2 * M2
0.005 L * M1 = 0.1 L * 0.835 mol/L
M1 = 16.7 mol/L or M Answer (c)
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