Question

1. If 1.000 gram of Na,coz is place in 0.5000 mL of water, what will be the pH of the solution? 2. If 1.000 gram of NaHCO, is place in 0.5000 mL of water, what will be the pH of the solution? 3. If 1.000 gram of H.CO, is place in 0.5000 mL of water, what will be the pH of the solution? 4. If 1.000 gram of Na CO, and 1.000 gram of NaHCO, are placed in 0.5000 mL of water, what will be the pH of the solution? 5. What will the equivalent point be if a 20.45 mL of a 0.1354 M solution of HCl is titrated with 0.1083 M base? 6. For the above problem, give the pH when 0, 4.000, 10.00 20.00, 30.00 and 35.00 mL of base is added. 7. What will the equivalent point be if a 19.38mL of a 0.1292 M solution of acetic acid is titrated with 0.1159 M base? 8. For the above problem, give the pH when 0, 4.000, 10.00, 20.00, 30.00 and 35.00 mL of base is added.

Answer 1

Solution 1: Giren Data Mass of NaCO3 = 1:00 g volume of waterz 0-500mL = 0.0005L Molar mass of Nag CO₂ = 2x23+ 12 +1663 i = 46+12+48 = 106 g/mol No of moles on of Naglą= Given mass Molecular mass = I g = 0.0094 mol 106 g/mol Scanned with CamScanner

Molarity = I no of moles of solute wolume of slection (L) Molarity of MacOz: 0.0094 mol 0.00052 = 18.67 M - Na coz 18.87M 2 Nat + ca ² - (2x 18 844 18.87M Hi + oH Co? +Ho - Ky = 2.1 x10-4 con +40 + 18.874 -x (18.84-xJM I Ec E HCO 0 tx t-on I - Intial 0 - Concentration tx C= Change in .. concentration (X)M E= Equilibrium Concoubatian. (X)M ] [4.07-1 Kb={HCO3] [ou [co? - J (Ho] Ko= x.x (18.87-x) assumed that 18.878) >X Scanned with CamScanner 8.8+8>>X So, 18.87-X = 18.8714 So, 18.87

Ro= x2 18.87 CMI * 1 Box 18.87 proved X= 2:1410-4x18.87 X=0.063M 18.87 M >> 0.063M Hence pon = -log[on] [OH) = x= 0.663M por= -log[o. 063M] por= 1.20 pH + por = pkw=14 PH= 14-por PH= 14-1-2 - 12.80 • Solietion 12.80. is Ics Scanned with CamScanner e RINSSON