A volume of 500.0 mL of 0.170 M NaOH is added to 525 mL of 0.200 M weak acid (K, -7.52 x 10-5). What is the pH of the resulting buffer? HA(aq) + OH-(aq) → H₂O(1) + A (aq) pH-
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- (a) Calculate the pH of the 0.30 M NH3 / 0.35 M NH4Cl buffer. What is the pH of the buffer after the addition of 0.030 mol HCl? note: Ka (NH4+) = 5.6 x 10 -10 NH3 (aq) + H+ (aq) → NH4+ (aq) (b) What are the hydronium [H3O+] and hydroxide [OH-] ion concentrations at 25°C in a 4.0 M aqueous Mg(OH)2.A volume of 500.0 mL of 0.170 M NaOH is added to 525 mL of 0.200 M weak acid (K, -7.52 x 10-5). What is the pH of the resulting buffer? HA(aq) + OH-(aq) → H₂O(1) + A (aq) pH-A 40.0-mL solution of 0.100 M ammonia (NH3; Kb = 1.8 x 10–5) is titrated with 0.100 M HCl solution. What is the pH of the solution after 20.0 mL of the acid solution is added? The net ionic equation is: NH3(aq) + H3O+(aq) --> NH4+(aq) + H2O (A) 9.55 (B) 9.25 (C) 8.20 (D) 5.28
- Ammonia is a weak base with a K₂ of 1.8 X 10-5. A 140.0-mL sample of a 0.200 M solution of aqueous ammonia is titrated with a 0.100 M solution of the strong acid HCI at 25 °C. The reaction is NH3 (aq) + HCI (aq) NH4+ (aq) Calculate the pH of the titration solution for each case. (a) Before any acid is added. (b) When the titration is at the half-equivalence point. (c) When the titration is at the equivalence point. (d) When the titration is 1.00 mL past the equivalence point. + CI (aq)1. Consider the titration of 25.0 mL of a weak acid, 0.100 M HCOOH (Ka = 1.8 x 10- 4) with 0.100 M NaOH. NaOH (aq) + HCOOH(aq) →H2O(1) + NaCHO2(aq) (i) What is the pH of the solution after adding 5.00 mL of NAOH? (ii) After adding 12.5 mL 2 Veq= midpoint pH = pKa = 3.74For the titration of 20.0 mL of 0.400 M HCl(aq) with 0.400 M NaOH(aq): 1. What is the initial pH of the HCl(aq)? HCl(aq) + NaOH(aq) NaCl(aq) + H₂O(l) pH = [Select] 2. What is the pH of the HCl(aq) after the addition of 10.00 mL of NaOH ? 3. What is the pH of the HCl(aq) after the addition of 20.0 mL of NaOH? pH = [Select] pH = [Select]
- Consider a flask containing 25.00 mL of 0.100M sodium benzoate (NaC, H5C00). This is titrated with a 0.110 M nitric acid solution from a burette. Given: K, of C,H;co0" is 1.6 x 10-10. Determine the pH of the solution in the flask after 32.00 mL of the acid has been added? You must show any reaction equation(s) that you may think are necessary. Determine the pH of the solution in the flask at the half-way point of the titration.A buffer solution contains 0.77 mol of hydrosulfuric acid (H2S) and 0.43 mol of sodium hydrogen sulfide (NaHS) in 5.70 L.The Ka of hydrosulfuric acid (H2S) is Ka = 9.5e-08.(a) What is the pH of this buffer?(b) What is the pH of the buffer after the addition of 0.44 mol of NaOH? (assume no volume change)(c) What is the pH of the original buffer after the addition of 0.08 mol of HI? (assume no volume change)A volume of 500.0 mL of 0.130 M NaOH is added to 605 mL of 0.200 M weak acid (Ka = 3.87 × 10-5). What is the pH of the resulting buffer? HA(aq) + OH(aq) · H,O(l) + A-(aq) pH = TOOLS V
- 5:50 1 Search Question 5 of 20 Submit Determine the mass of solid NaCH;COO that must be dissolved in an existing 500.0 mL solution of 0.200 M CH3COOH to form a buffer with a pH equal to 5.00. The value of Ka for CH-COОH is 1.8 х 10-5. 1 2 Let x represent the original concentration of CH;COO in the water. Based on the given values, set up the ICE table in order to determine the unknown. CH3COOH+ H20(1) =H;O*(aq) +CH3COO-(a Initial (M) Change (M) Equilibrium (M) 5 RESET 0.200 5.00 -5.00 1.0 x 10-9 -1.0 × 10-9 1.0 x 10-5 -1.0 x 10-5 1.8 x 10-5 -1.8 x 10-5 х+ 5.00 x - 5.00 x + 1.0 × 10-9 х - 1.0 х 10-9 1.0 x 10-5 x - 1.0 × 10-5 x + 1.8 × 10-5 х - 1.8 х 10-5How much 5.90 M NaOH must be added to 450.0 mL of a buffer that is 0.0215 M acetic acid and 0.0270 M sodium acetate to raise the pH to 5.75? CH3COOH(aq)+OH-(aq)--->CH3COO-(aq)+H2O(l)culate the pH of a buffer solution 1) Calculate the pH of a solution prepared by dissolving 2.05 g of sodium acetate, CH3COONa, in 85.0 mL of 0.10 Macetic Assume the volume change upon dissolving the sodium acetate is negligible. K₂ of CH3COOH is acid, CH3COOH(aq). 1.75 x 10-5. pH =