pH = -log[H+]
HCl -> H+ + Cl-
KCl -> K+ + Cl-
HCl releases H+ but KCl does not release H+ neither OH- , so it does not add to the acidity or basicity of the solution.
Therefore we will calculate pH only due to HCl.
HCl(0.01M) -> H+(0.01M) + Cl-(0.01M)
Since H+ concentration is 0.01 , therefore pH = -log[0.01]=2
Using activities, calculate the pH of the mixture: 0.01 M HCI and 0.04 M KCI.
Use activities to calculate the molar solubility of Zn(OH)_2 (Ksp=3 times 10^-16) in 0.01 M KCI.
27. Calculate the pH of a mixture made by adding 50.0 mL of 0.20 M HCI(aq) to 150.0 mL of water at 25.0°C A) 0.70 B) 1.00 C) 1.18 D) 1.30 E) 13.0 Ans: D 28. A mixture is made by adding 50.0 mL of 0.20 M NaOH(aq) to 50.0 mL of water. At 25.0°C, what is its pH? A) 1.00 B) 4.55 C) 7.00 D) 13.0 E) 13.3
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can you help me with these 1. Calculate the pH of a 0.01 mM (1.0 mM = 10-3M) solution of HCI? 2. Calculate the pH of a 0.02 M solution of NaOH.
Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.31-M HCI to 620. mL of each of the following solutions Change is defined as final minus initial, so if the pH drops upon mixing the change is negative a) water pH before mixing 7.00 pH after mixing- 1.42 pH change = X 5.53 1- b) 0.186 M C2H302 pH before mixing- 9.01 pH after mixing 5.34...