What is the OH- concentration of a 0.08M solution of CH3COONa [(KaCH3COOH)=1.8*10^-5]
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pH=7+1/2pKa+1/2log C (formula of pH of salt of weak acid and strong base)
Here, CH3COONa is the salt of acetic acid and sodium hydroxide.
Therefore, pOH=14-pH
=>-log OH=14-pH
=>log OH=pH-14
=>OH=10^(pH-14)=10^(0.5pKa+0.5logC-7)=10^(-0.5log1.8+2.5+0.5log8-1-7)=10^(-0.13+2.5+0.45-8)=10^(-5.18)
Here, CH3COONa is the salt of acetic acid and sodium hydroxide.
Therefore, pOH=14-pH
=>-log OH=14-pH
=>log OH=pH-14
=>OH=10^(pH-14)=10^(0.5pKa+0.5logC-7)=10^(-0.5log1.8+2.5+0.5log8-1-7)=10^(-0.13+2.5+0.45-8)=10^(-5.18)
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