The pH of a 0.001 mol dm\(^{-3}\) of  H\(_2\)SO\(_4\) is 

[Log\(_{10}\)2 = 0.3]

daisyxlove

3 months ago

H2SO4 - 2H +SO4
pH = -log [H+]
H = 0.001 mol/dm
pH = - log [ 2 × 0.001]
pH = - log [ 2 × 10^-3]
n - log a
n= 3
loga =log 2 = 0.3
3 - 0.3 = 2.7

OladipupoDaniel1233

2 months ago

I choosed 2.7 you choosed another thing for me

OladipupoDaniel1233

2 months ago

This answer u picked for me is not correct cross-check now

OladipupoDaniel1233

2 months ago

This answer u picked for me is not correct cross-check now make amend