Question

1. What volume of 0.10 M NaOH is needed to neutralize 100 mL of 0.050 M HCl?

2a. When 100 mL of 0.01 M NaOH solution is mixed with 100 mL of 0.01 M of HNO₃ solution, what is the concentration of H⁺ and the pH in the resulting mixture?

2b. When 100 mL of 0.01 M NaOH solution is mixed with 100 mL of 0.02 M of HCl solution, what is the concentration of H⁺ and the pH in the resulting mixture?

2c. When 100 mL of 0.01 M Mg(OH)₂ solution is mixed with 100 mL of 0.01 M of HCl solution, what is the concentration of H⁺ and the pH in the resulting mixture?'

2d. A 100 mL solution of 0.010 M CH₃COOH (acetic acid) has a pH = 3.4 and [H⁺] = 4.0 × 10⁻⁴ M. How much base is needed to react with all the acetic acid, and what is the pH of the resulting mixture?

Answer 1

There all questions can be solved coith 120 the help of simple Equation - milli Egviral out of good = mit Equinolent y Rose NIV = Na V2 ay 0.10x one = loox 0.050 VNoon (needed) = 50 mL Assuming Mg (04) as a strory sast or base One mole rig (04) caill give two mole 340, keeping this in mind apply the same Normality Equation (in terme cy molar) - Concentration). the miliegriaut af one left - - 100x (6-0) X 2) = 100% ovoj ( HCI My fo 42 Sinu both aard and base uret are strory and the miliequivalents used is also same, so Complete neutry lization Coill take plan and the 17 be i lu+7=107 pu=7 take plan due to and the 4t qrailable will auto protolpiscy agtet. - 100x oiol (midi Erri ef and left) total volume of solution has now become 200 me so - 100 x0.ol 10o xoool = 0.005 Con-] = after addition N904 / 100ml ,0.01) in (100 mL, 0.02 19 4 ) totul volum will become goom and in this volume onto the sicmainly milieuilent ay nas iconxun oorool) = (10080.0l) C O2 (mxv) So the molasty cy Now Solun - MXV - rorool 200 200 Tht = 0.05 mollis z 0.005 pon - 2.30 ph= 14-pon = 14-2.3 = 11.7 ph=- lug [o.oos] = 2.30