Question

В. Titration of KHP Trial 1 Trial 2 Trial 3 Trial 4 0.321ps 0.00 mL 20mL Q302. 0.334 0.00 mL .10 mL v.O mL 0.328 O.00 mL Ke.20 mL 0.20 mL 1. Mass of KHP T-00 mL Initial buret reading 2. 510 5 10 mL Final buret reading 3. mL .20 ml. 4. Volume NaOH used Moles KHP used 5. mol mol mol mol 6. Moles NaOH used mol mol mol mol 7 Molarity of NaOH M. М М 8. Average Molarity Calculations: 1510 I.10 -20 10.20 - 000 -0.00-0C0 00 5.10 .10 10. 20 M M

Answer 1

We have , No. of moles = Mass / Molar mass

$\therefore$ No. of moles of KHP = Mass of KHP / Molar mass of KHP

Molar mass of KHP = 204.22 g/mol

	Trail 1	Trail 2	Trail 3	Trail 4
No. of moles of KHP	= 0.326 g / 204.22 g/mol = 0.001596 mol	= 0.302 g / 204.22 g/mol =0.001479 mol	= 0.334 g / 204.22 g/mol = 0.001635 mol	= 0.328 g /204.22 g/mol = 0.001606 mol

Consider reaction KHP + NaOH KNaP + H₂O

From reaction, 1 mole KHP reacts with 1 mole NaOH. Therefore, we can write

No of moles of KHP = No of moles of NaOH

Therefore, we get no .of moles of KOH equal to no. of moles of KHP

	Trail 1	Trail 2	Trail 3	Trail 4
No. of moles of NaOH	0.001596 mol	0.001479 mol	0.001635 mol	0.001606 mol

We know that, Molarity = No of moles / volume of solution in L

	Trail 1	Trail 2	Trail 3	Trail 4
Volume of NaOH in L	16.20 ml = 0.0162 L	15.10 ml = 0.0151 L	16.10 ml = 0.0161 L	16.20 ml = 0.0162 L
[NaOH]	= 0.001596 mol / 0.0162 L = 0.0985 M	= 0.001479 mol/0.0151 L = 0.0979 M	= 0.001635 mol / 0.0161 L = 0.102 M	= 0.001606 mol / 0.0162 L = 0.0991 M
Average [NaOH]		= ( 0.0985 + 0.0979 + 0.102 + 0.0991 ) / 4 = 0.0994 M