mohon bantuan jawabannya

<p><strong><u>Jawabannya adalah D.</u></strong></p><p>&nbsp;</p><p>Campuran basa lemah NH4OH dan asam kuat HCl yang habis bereaksi maka akan membentuk larutan penyangga basa.</p><p>nHCl = M.V = 0,1 x 50 = 5 mmol</p><p>nNH4OH = M.V = 0,1 x 100 = 10 mmol</p><p>&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; NH4OH &nbsp; &nbsp;+ HCl &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;---&gt; NH4Cl &nbsp; &nbsp;+ &nbsp; &nbsp; H2O</p><p>M &nbsp; &nbsp;10 mmol &nbsp; &nbsp; 5 mmol &nbsp;</p><p>R &nbsp; &nbsp;-5 mmol &nbsp; &nbsp; &nbsp;-5 mmol &nbsp; &nbsp; &nbsp; &nbsp;+ 5 mmol</p><p>S &nbsp; &nbsp; &nbsp;5 mmol &nbsp; &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 5 mmol</p><p>Bersisa 5 mmol NH4OH dan terbentuk 5 mmol NH4Cl</p><p>Perhitungan pH campuran</p><p>[OH-] = Kb.nNH4OH/nNH4Cl.valensi</p><p>= 10^-5 x (5 mmol/5mmol.1)</p><p>= 10^-5 M</p><p>pOH = -log [OH-]</p><p>= -log 10^-5</p><p>= 5</p><p>pH = 14 - pOH</p><p>pH = 14 - 5 = 9</p><p>&nbsp;</p><p><strong>Jadi, pH campuran sebesar 9.</strong></p>