Hitunglah pH larutan 1,64 gram natrium asetat (CH3COONa) dalam 200 mL larutan (Ka CH3COOH =10^-5, Ar Ca = 12, H = 1, O = 16, Na = 23)

<p>Jawaban yang benar adalah 9</p><p>&nbsp;</p><p>Pembahasan</p><p>Hidrolisis adalah penguaraian garam oleh air membentuk asam dan basanya kembali</p><p>Garam yang dapat terhidrolisis dalam air adalah garam yang tersusun dari asam lemah atau basa lemah</p><p>natrium asetat (CH3COONa) merupakan garam dari asam lemah dan basa kuat, maka dapat terhidrolisis sebagian</p><p>Persamaannya sebagai berikut</p><p>CH3COONa &nbsp; → &nbsp; CH3COO^- &nbsp; + &nbsp; Na⁺</p><p>CH3COO^- &nbsp;bereaksi dengan air</p><p>CH3COO^- &nbsp;+ &nbsp; H2O &nbsp; ⇄ &nbsp; CH3COOH &nbsp; + &nbsp; OH^-</p><p>Na⁺ tidak bereaksi dengan air&nbsp;</p><p>Maka [OH-] dapat dicari dengan</p><p>[OH-] = √(Kw/Ka x [CH3COO^- ]</p><p>&nbsp;</p><p>Konsentrasi CH3COONa = (massa/Mr) x (1000/V)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = (1,64 gr / 82 gr/mol) x (1000/200 mL)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 0,02 x 5</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 0,1 M</p><p>[CH3COONa] = [CH3COO^-] = 0,1 M = 10^-1 M</p><p>&nbsp;</p><p>[OH-] = √(Kw/Ka x [CH3COO^- ]</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = √(10^-14 / 10^-5 x 10^-1 M</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = √10^-10&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 10^-5 M</p><p>&nbsp;</p><p>pOH = -log 10^-5 M</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 5</p><p>pH = 14 - pOH<br>&nbsp; &nbsp; &nbsp; &nbsp; = 14 - 5</p><p>&nbsp; &nbsp; &nbsp; &nbsp; = 9</p><p>&nbsp;</p><p>Maka larutan tersebut adalah 9</p>