Answers: 2
![[base]=0.28M](/tpl/images/0733/9118/c67fa.png)
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](/tpl/images/0733/9118/b44a5.png)
![[base]=1.38[acid]=1.38*0.20M=0.28M](/tpl/images/0733/9118/1652c.png)
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