Question

Calculate and enter the molarity of your three HCl standardization trials using the volume of standardized NaOH solution required for each and the average molarity of the NaOH solution from the standardization trials with KHP. You should report 3 significant figures, e.g. 0.488 M.

Scoring Scheme: 3-3-2-1. Calculate and enter the molarity of your three HCI standardization trials using the volume of standardized NaOH solution required for each and the average molarity of the NaOH solution from the standardization trials with KHP. You should report 3 significant figures, e.g. 0.488 M. Entry # MHCI Vol HCI(mL) 10.00 10.00 Vol NaOH(mL) 21.50 #1: #2: 21.30 #3: 10.00 21.43

I need to find the average molarity of my NaOH to solve this, but I am unsure on how to solve for it. It had a concentration of 0.1 M. Thank you!

Answer 1

Answer :

Reaction is :

HCl + NaOH -----> NaCl + H_{​​​​​​2}​​​​O

1 mole of NaOH is required to completely neutralise 1 mole of HCl .

At equivalent point : no. of moles of HCl = no .of moles of NaOH

no. of moles = Molarity (M) * Volume (V in litres)

So, (MV)_HCl = (MV) _NaOH

_{​​​​​​}M_{​​​​​HCl}  = (MV) NaOH / V_{​​​​HCl}

use this formula to calculate Molarity of HCl solution ( given that Molarity of NaOH solution is 0.1 M

Entry	volume of HCl solution (ml)	volume of NaOH solution (ml)	Molarity of HCl solution (M)
1	10	21.5	= 0.1 * 21.5 / 10 = 0.215 M
2	10	21.3	=0.1 * 21.3 / 10 = 0.213 M
3	10	21.43	=0.1 * 21.43 / 10 = 0.2143 M

Average Molarity of HCl solution = (0.215 + 0.213 + 0.2143 ) / 3 = 0.6423 / 3 = 0.2141 M

In 3 significant figures answer will be 0.214 M