500.0 mL of 0.170 M NaOH is added to 615 mL of 0.250 M weak acid (Ka = 1.08 × 10-5). What is the pH of the resulting buffer?
inal concentration of NaOH = 0.17*500/(500+615) = 0.07623 M ( =
concentration base)
final concentration of acid = 0.25*615/(500+615) = 0.1379 M -
0.07623 M = 0.06166
pH = -log(1.08*10^-5) + log(0.07623/0.06166)
pH = 4.97 + 0.0921 =5.05
500.0 mL of 0.140 M NaOH is added to 535 mL of 0.250 M weak acid (Ka = 4.06 × 10-5). What is the pH of the resulting buffer?
500.0 mL of 0.100 M NaOH is added to 535 mL of 0.250 M weak acid (Ka = 7.69 × 10-5). What is the pH of the resulting buffer?
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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 =
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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)
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 =