chemistry

Calculate the change in pH when 9.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

Calculate the change in pH when 9.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

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  1. pOH = pKb + log [NH4+]/ [NH3]

    pKb of ammonia = 4.74

    initial pOH = 4.74 + log 0.100/0.100 = 4.74
    pH = 14 - 4.74=9.26

    moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100
    moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300

    NH3 + H+ = NH4+
    moles NH3 = 0.0100 - 0.000300=0.00970
    moles NH4+ = 0.0100 + 0.000300=0.0103

    pOH = 4.74 + log 0.0103/ 0.00970= 4.77
    oH = 14 - 4.77 = 9.23

    delta pH = 9.26 - 9.23 =0.03

    adding OH- the reaction is
    NH4+ + OH- = NH3 + H2O
    the moles of NH4+ will decrease and the moles of NH3 will increase
    I can not answer your second question because thr molar concentration of NaOH is not given

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