anonymous
  • anonymous
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).
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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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