Answers: 1
![[LiOH]=\frac{0.00225mol}{0.1750L}=0.013M](/tpl/images/0917/1926/d60d2.png)
![pOH=-log([OH^-])=-log(0.013M)=1.89](/tpl/images/0917/1926/5deb1.png)
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A 100.0 mL sample of 0.18 M HClO4 (Strong acid) is titrated with 0.27 M LiOH. Determine the pH of th...
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