A 100. mL sample of 0.10 M HCl is mixed with 50. mL of 0.14 MNH...

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Answer 1

Answer #1

moles of HCl = molarity×volume/1000

= 100×0.10/1000 = 0.01 moles

Moles of NH3 = 50×0.14/1000 = 0.007 Moles

Moles of HCl left after consuming NH3 = 0.01-0.007

= 0.003 moles

Total volume = 100+50 = 150 ml

[H+] = moles×1000/volume = 0.003×1000/150 = 0.02

pH = - log[H+] = - log(0.02) = 1.70

Answer is d).

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Answer 2

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