Answer:
a.
[tex][H _{3}O {}^{ + } ] = 4.23 \times {10}^{ - 3} M[/tex]
b.
[tex][OH {}^{ - } ] = \frac{kw}{[H {}^{ + } ]} \\ [OH {}^{ - } ] = \frac{1 \times {10}^{ - 14} }{4.23 \times {10}^{ - 3} } \\ [OH {}^{ - } ] = 2.36 \times {10}^{ - 12} M[/tex]
c.
[tex]pH = - log[H {}^{ + } ] \\ pH = - log(4.23 \times {10}^{ - 3} ) \\ pH = 2.37[/tex]
d.
[tex]pOH = - log[OH {}^{ - } ] \\ = - log(2.36 \times {10}^{ - 12} ) \\ = 11.63[/tex]
