Part S ACID BASE TITRATION (neutralization experiment): (for this experiment,I d
ID: 573528 • Letter: P
Question
Part S ACID BASE TITRATION (neutralization experiment): (for this experiment,I drop-0.05 mL) 17. Get a dropper of unknown molarity HNO, and a dropper of 0.10 M NaOH. Add 10 drops of acid to the beaker and add a few drops of universal indicator. Now add drops of 0.10 M NaOH until the acid is neutralized. USE THIS FORMULA #11 x M.ius v, "apoi 1 x M.-x v.." a) How many ml. of base were needed to neutralize the 10 drops of acid? b) Calculate the molarity of the HNO c) Practice: If 25 mL of 0.10 M NaOH neutralize 15 mL of H,SO, what is the molarity of the H,SO.? ze l 5 mL of HNO, what is the molanty oftheHNO.. d) Practice: If 15 mL of 0.10 M e) Practice: If 1S ml. of 0, 10 M NaOH neutralize 25 ml. of HCI, what is the molarity of the HC1? Solution ,5mlO 2. Part 6-COLLIGAT e for homework . Look at the results in the data table below and answer question W2 below 6 min 140 man Time required to Boil Time required to Freeze 5 unin 120 min mmin 160 minExplanation / Answer
Ans. #17.b Given-
Volume of NaOH solution consumed = 0.8 mL
Molarity of NaOH solution = 0.10 M
Volume of HNO3 solution taken = 10 drops
= 10 drops x (0.05 mL/ drop) = 0.50 mL
# Method 1: Balanced reaction: HNO3(aq) +NaOH(aq) -------> NaNO3(aq) + H2O(l)
According to the stoichiometry of balanced reaction, 1 molHNO3 is neutralized by 1 mol NaOH.
So, at equivalence point, the moles of NaOH consumed must be equal to the moles of HNO3 taken.
Now,
Moles of NaOH consumed = Molarity of NaOH x Volume of NaOH solution in liters
= 0.10 M x 0.8 mL ; [1 mL = 0.001 L]
= (0.10 mol/ L) x 0.0008 L ; [1 M = 1 mol/ L]
= 0.00008 mol
So, moles of HNO3 taken = Moles of NaOH consumed = 0.00008 mol
# Volume of HNO3 solution taken = 0.50 mL = 0.0005 L
Now,
Molarity of HNO3 = Moles of HNO3 / Volume of HNO3 solution in liters
= 0.00008 mol / 0.0005 L
= 0.16 mol/ L
= 0.16 M
# Method 2: At equivalence point-
A x (M1V1), acid = B x (M2V2), base
Where, A = number of H+ given by acid
B = number of OH- given by base
M = Molarity of respective solutions
V = Volume of respective solution
Here, both the acid and bases are monoprotic and monobasic. So, the values of A=B=1.
Given-
Volume of NaOH solution consumed = 0.8 mL
Molarity of NaOH solution = 0.10 M
Volume of HNO3 solution taken = 10 drops
= 10 drops x (0.05 mL/ drop) = 0.50 mL
Putting the values in above equation-
1 x (M1 x 0.50 mL) = 1 x (0.10 M x 0.80 mL)
Or, M1 = (0.10 M x 0.80 mL) / 0.50 mL
Hence, M1 = 0.16 M
Therefore, Molarity of HNO3 = 0.16 M