The original given concentration was 25.0mL of 0.100 M HCO2H (formic acid Ka= 1.8x10^-4) with 0.100 M of NaOH.
The initial [HCO2-] = 0 M
The change in [HCO2-] = 0.0025 mol/{(25+25)/1000 L} = 0.05 M
The equilibrium [HCO2-] = 0 + 0.05 = 0.05 M
pH = 7 + 1/2 (pKa - Log[HCO2-]) = 7 + 1/2 {-Log(1.8*10-4) - Log(0.05)} = 9.52
The original given concentration was 25.0mL of 0.100 M HCO2H (formic acid Ka= 1.8x10^-4) with 0.100...
Formic acid (HCO2H) has a Ka value of 1.70 X 10-4 at 25°C. Calculate the pH at 25°C of . . . . a. a solution formed by adding 15.0 g of formic acid and 30.0 g of sodium formate (NaCO2H) to enough water to form 0.500 L of solution. b. a solution formed by mixing 30.0 mL of 0.250 M HCO2H and 25.0 mL of 0.200 M NaCO2H and diluting the total volume to 250 mL. c. a solution...
2. Consider the titration of 25.0mL of 0.100M formic acid with 0.100 M NaOH (Ka= 1.8 x 10-4). Part IV. What is the pH after adding 25.0mL of NaOH? a)5.60 b) 8.23 c) 9.80 d)13.00 Answer is B
Concentration of HCI is 0.09655 M. Concentration of NaOH is 0.100 M. a) What is the pH at equivalence point? b) Moles of HCl at equivalence point? c) Moles of NaOH at equivalence point? d) Volume of NaOH at equivalence point? Volume NaOH added (mL) pH Moles of NaOH added 0.89 0.89 0.00010 0.89 0.00020 0.90 0.00030 0.91 0.00040 0.93 0.00050 0.94 0.00060 0.96 0.00070 0.98 0.00080 1.01 0.00090 1.03 0.00100 1.06 0.00110 LOS 0.00120 1.11 0.00130 1.15 0.00140 1.19...
A 25.0 ml sample of 0.20 M Formic Acid (HCO2H, aq) is titrated with 0.10 M KOH(aq). What is the pH after 50.0 mL of the 0.10 M KOH has been added?
1. Calculate pH and % ionization of 0.25 M Naco (Ka (HCO2H) 1.8 x 10") 2- Calculate the pH of: a) 0.35 M HNO b) 0.15 M Sr(O)2 c) 0.08 M Ba(CIO4)2 3- Calculate the pH of 0.375 L buffer solution made of a 0.18 M Acetic acid HC2H:02(Ka 1.8x 10) and a 0.134 M Potassium acetate KC2Hs02 (do not use more than 3 digits beyond the decimal point) a) before adding anything b) after adding 0.010 mol Ba(OH)2 c)...
Concentration of HCl is 0.09655 M. Concentration of NaOH is 0.100 M. a) What is the pH at equivalence point? b) Moles of HCl at equivalence point? c) Moles of NaOH at equivalence point? d) Volume of NaOH at equivalence point? 1.15 Volume NaOH added (mL) pH Moles of NaOH added 0.89 0.89 0.00010 0.89 0.00020 0.90 0.00030 0.91 0.00040 0.93 0.00050 0.94 0.00060 0.96 0.00070 0.98 0.00080 1.01 0.00090 1.03 0.00100 1.06 0.00110 1.08 0.00120 1.11 0.00130 0.00140 1.19...
Consider the titration of the titration of 50.0 mL of 0.100 M acetic acid (HC2H2O2) with 0.100 M. The pka = 4.76. d. Determine the pH after 50.0 mL of titrant (NaOH) have been added. This is the equivalence point. All of the acid has been converted to its conjugate base, pH is determined by the equilibrium for the conjugate base
You are provided with 100 mL of 0.100 M acetic acid (CH3COOH, Ka = 1.8 x 10-5), which you will be asked to titrate with 0.050 M sodium hydroxide (NaOH). (a) What is the pH after the addition of 80 mL of the 0.050 M sodium hydroxide solution? Show your work. In your answer, show the reaction that occurs when the sodium hydroxide is added. (b) What is the pH at of the solution at the equivalence point (where the...
Ka for hypochlorous acid. HCIO, is 3.0x108. Calculate the pH after the stepwise addition of 0.100 M NaOH (from 0 ml to 50 mL, in I mL increments) to 40.0 ml of 0.100 M HCIO. Plot your results in Excel and on the graph, identify the pH at the equivalence point and at half of the equivalence point.
500.0 mL of 0.100 M NaOH is added to 595 mL of 0.250 M weak acid (Kg = 4.90 105). What is the pH of the resulting buffer? HA(aq) + OH(aq) → H,O(l) + A+ (aq) Number pH =