1000[tex] cm^{3} [/tex] of HClO_{4} [/tex] contain 0.00046mol
16[tex] cm^{3} [/tex] of HClO_{4} [/tex] contain x
x = [tex] \frac{16cm ^{3} * 0.00046mol}{1000cm x^{3} } [/tex]
x = 0.00000736mol
[tex]M_{1} V_{1} = M_{2} V_{2} [/tex]
(0.00000736mol)(16[tex] cm^{3} [/tex])= [tex](x) (42cm^{2})
[/tex]
[tex]x= \frac{(0.00000736mol) * (16cm^{3}) }{42cm^{3} } [/tex]
= 0.000002803mol
pH= -log [[tex] H^{+} [/tex]]
= -log (0.000002803cm[tex] ^{3} [/tex]
= 5.55
