A10.0 ml solution of 0.300 m nh3 is titrated with a 0.100 m hcl solution. calculate the ph after the addition of 10.0 ml hcl.

pH = 9.55

[NH3] = 2.0 mmol/20.0 mL = 0.100 M

[NH4+] = 1.00 mmol/20.0 mL = 0.0500 M

Ka = (1.0 × 10^-14)/(1.8 × 10^-5)

     = 5.56 × 10^-10

pKa = log Ka = 9.25

pH = pKa + log ([base]/[acid])

    = 9.25 + log (0.100/0.0500)

    = 9.55

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