Pke cara yaa

<p>Jawabannya : D. 50 ml CH3COOH &nbsp;0.2 M + 50 ml NaOH 0.1 M</p><p>Untuk pembahasan opsi lain, kenapa bukan larutan &nbsp;penyangga bisa kamu buktikan dengan mereaksikannya.&nbsp;</p><p>Untuk larutan penyangga sisa yang di dapat harus berupa asam/basa lemah dan garamnya dengan asam/basa kuat harus habis bereaksi.</p><p>Ini persamaan reaksi opsi D.</p><p>Asam lemah + &nbsp;Basa kuat -&gt; garam</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;CH3COOH &nbsp; &nbsp;+ &nbsp; &nbsp; &nbsp; &nbsp; NaOH &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;---&gt; &nbsp; &nbsp; CH3COONa</p><p>(mula²) &nbsp; &nbsp; &nbsp;M= n x v &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;M= n x v</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 10 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 5 mmol</p><p>(Bereaksi) 5 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 5 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;5 mmol</p><p>(Sisa) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 5 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;_ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;5 mmol</p><p>Jadinya larutan penyangga basa</p><p>Nah, simple nya larutan penyangga itu intinya yang harus bersisa adalah Asam/basa lemah dan garamnya.</p><p>* Ingat asam kuat tidak bisa bereaksi dengan basa kuat; begitupun sebaliknya.</p><p>&nbsp;</p>