A 10.0 mL solution of 0.540 M NH3 is titrated with a 0.180 M HCl solution. Calculate the pH after the following additions of the HCl solution: 0.00 mL, 10.0 mL, 30.0 mL, and 40.0 mL. Round your answer to 3 decimal places. Note: Reference the K of acids at 25 °C and K, of bases at 25 °C tables for additional information. Part 1 of 4 0.00 mL Added, pH = 11.493 Part 2 of 4 10.0 mL Added, pH = 9.556 x10 Part 3 of 4 30.0 mL Added, pH = 9.255 Part 4 of 4 40.0 mL Added, pH = 1.444 x10 X x x G G G G

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Chapter15: Acid-base Equilibria
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A 10.0 mL solution of 0.540 M NH3 is titrated with a 0.180 M HCl solution. Calculate the pH after the following additions
of the HCl solution: 0.00 mL, 10.0 mL, 30.0 mL, and 40.0 mL. Round your answer to 3 decimal places.
Note: Reference the K of acids at 25 °C and K, of bases at 25 °C tables for additional information.
Part 1 of 4
0.00 mL Added, pH = 11.493
Part 2 of 4
10.0 mL Added, pH = 9.556
x10
Part 3 of 4
30.0 mL Added, pH = 9.255
Part 4 of 4
40.0 mL Added, pH = 1.444
x10
X
x
x
G
G
G
G
Transcribed Image Text:A 10.0 mL solution of 0.540 M NH3 is titrated with a 0.180 M HCl solution. Calculate the pH after the following additions of the HCl solution: 0.00 mL, 10.0 mL, 30.0 mL, and 40.0 mL. Round your answer to 3 decimal places. Note: Reference the K of acids at 25 °C and K, of bases at 25 °C tables for additional information. Part 1 of 4 0.00 mL Added, pH = 11.493 Part 2 of 4 10.0 mL Added, pH = 9.556 x10 Part 3 of 4 30.0 mL Added, pH = 9.255 Part 4 of 4 40.0 mL Added, pH = 1.444 x10 X x x G G G G
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