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Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8*10^-5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.

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A 30.0-ml volume of 0.50 M CH3COOH (Ka=1.8*10^-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.

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Please put your subject in the School Subject space, so someone with expertise will answer your question. This is not my area of expertise.

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Both of these problems have moles acid = moles base which means you are at the equivalence point of each. Therefore, the pH is determined by the pH of the salt.

Write the hydrolysis equation, set up an ICE chart, and solve for (H^+), then convert to pH. One important point to remember (and easy to forget) is that the concn of the salt form is moles acid or base/TOTAL volume at the equivalence point.

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