Question

Standardization of a Sodium Hydroxide Solution Section Name Score Date Post-lab Assignment 1. Calculate the molarity of a barium hydroxide solution if you used 41.65 mL of it to neutralize 1.190 g of potassium hydrogen phthalate. Ba(OH)2(aq) + 2 KHC4H6O4(aq) + BaC,H,O4(aq) + K2CH4O4(aq) + 2 H300) 2. Write a balanced chemical equation for the reaction of hydrochloric acid with sodium hydroxide. 3. Calculate the molarity of a hydrochloric acid solution if 34.21 mL of 0.0431M sodium hydroxide neutralizes 25.00 mL of the acid solution. Write a balanced chemical equation for the reaction of phosphoric acid with sodium hydroxide. Calculate the molarity of a phosphoric acid solution if 34.21 mL of 0.0431M sodium hydroxide neutralizes 25.00 mL of the acid solution.

Answer 1

Solution: Griven Balance Reaction:- Ba(OH)2 (aq) + 2 KMC8H4041aq) → BaC2H494 (aq) + K2 C2H4O4CxH) - 2H2011) mass of potassium hyclrogen phthalate = 1.1909 ar mass of Potassium hydrogen pthalate- 204.23 g/mol So mole of Potassium hydroge phthalate: 1.1908 - 25.83X10-3M01 204.23g /mol According to beuqyxe Reaction - 2 mol KHP (Potassium hydrogen pht HP (Potassium hydrogen phthalate) React with 1 mol [139 (0H), 7 So I mol KHP React with Imol Ba(OH)2 So• 5683X10-3 mol KHP Recict with 145.83 X10-3 mol BacOH)2 Mole of Ba(OH), = 2.91.X10-3 mol Volume=4765me or 41.65X10-3L i L=1000 m , Mol wity Mole volume (1) Then molarity of Ba(014), = 2.9 1.X1023 mol 41.65810-3 L 20.0699 M Ang where M, = 0.0431M Cu 2 Balance chemical Reaction HCI + NaOH + NaCl + H2O CW 3 we know that at equivalence point · MiV = M2 V2 son M2 = MN V = 34.21 me Me=? V = 25m2 + 0.0431MX3 4.21 mat 25 mclarity of HCl = 0.0589 M Angs phosphuric acid (M3 P04) and Sodium Hy dieroxide (NOOM) H3PO4 +3N90H N93 P04 + 3H30 ou.4' Phosphoric acid so

ou 5. Given that's molarity of NGOH = 0.0431M ; Volume = 34.21ml or 34.21410-3 camoles of NaOH= 0.0431moly 34:21x10-3Liker =) 1.48X10-3 mol Liter woording to balance chemical Reaction (aue. 4) 3 mol NaOH React with 1 more H₂PO4 Amol ; volume = 25 mi or 25416-3, So I mol N90H React with {mol H3PO4 sex 103 mol NCOH React with $ x 1.48X10-3 mol H2PO4 co mole of H3 P04 = 4:93410-4 mol ; vcheme = 25 min so molarity of H3PO4 = 4.93810-4 mol 25x10-31 molarity of H3 PO4 =0.01973 M A