10 mL x 0.1M HCl = 1 mmols.
72.5 mL x 0.2M = 14.5 mmols NH3.
72.5 mL x 0.25M = 18.125 mmols NH4^+. You can round to th correct number of significant figures when you finish the calculation.
.........NH3 + H^+ ==> NH4^+
I........14.5...0.......18.125
add.............1.............
C.......-1.....-1........+1
E.......13.5....0........19.125
pH = pKa + log(13.5/19.125) = ?
Calculate the pH of the 0.20 M NH3/0.25 M NH4Cl buffer.
I calculated the pH to be 9.15
What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 72.5 mL of the buffer?
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