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(1). 500.0 mL of 0.110 M NaOH is added to 535 mL of 0.250 M weak acid (Ka = 6.08 × 10-5). What is the pH of the resulting buffer? HA(aq) + OH-(aq) = H2O(l) + A- (aq) (2). A 25.863 g sample of aqueous waste leaving a fertilizer )manufacturer contains ammonia. The sample is diluted with 74.612 g of water. A 14.159 g aliquot of this solution is then titrated with 0.1059 M HCl. It required 31.31 mL of...
A volume of 500.0 mL of 0.140 M NaOH is added to 585 mL of 0.250 M weak acid (K, = 1.73 x 10-). What is the pH of the resulting buffer? HA(aq) + OH(aq) — H, 0(1) + A- (aq) pH =
A volume of 500.0 mL of 0.180 M NaOH is added to 565 mL of 0.250 M weak acid (Ka= 4.73 x 10-5) What is the pH of the resulting buffer? HA(aq) + OH-(aq) -> H2O(l) + A-(aq)
A volume of 500.0 mL of 0.120 M NaOH is added to 605 mL of 0.250 M weak acid (Ka=3.72×10−5). What is the pH of the resulting buffer? HA(aq)+OH-(aq)----->H2O(l)+A-(aq)
A volume of 500.0 mL of 0.160 M NaOH is added to 625 mL of 0.250 M weak acid (Ka = 5.24 x 10-5). What is the pH of the resulting buffer? HA(aq) + OH(aq) + H2O(1) + A- (aq) pH =
A volume of 500.0 mL of 0.110 M NaOH is added to 625 mL of 0.200 M weak acid (K, = 2.11 x 10"). What is the pH of the resulting buffer? HA(aq) + OH(aq) + H2O(l) + (aq)
A volume of 500.0 mL of 0.150 M NaOH is added to 525 mL of 0.250 M weak acid (?a=4.43×10−5). What is the pH of the resulting buffer? HA(aq)+OH−(aq)⟶H2O(l)+A−(aq) pH=
A volume of 500.0 mL of 0.120 M NaOH is added to 605 mL of 0.250 M weak acid (K, = 7.89 x 10-). What is the pH of the resulting buffer? HA(aq) + OH(aq) → H2O(l) + A™(aq) pH II
A volume of 500.0 mL of 0.140 M NaOH is added to 555 mL of 0.250 M weak acid (Ka-5.49 x 10-5). what is the pH of the resulting buffer? HA(aq) +OH(aq)H,0) + A (aq) PH-
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 =