How many milliliters of concentrated HCl (12.2 M) should be diluted to prepare 200.0 ml of an acid solution with a pH of 1.50?
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First, we need to find the number of moles of #H^+# ions in #"200 mL"# (#"0.200 L"#) of an acid with #"pH 1.50"#.
To do this, we can find the concentration of #H^+# and multiply that by #"0.200 L"#:
#[H^+] = 10^(-1.50) = "0.0316 M"#, or #"0.0316 mol/L"#
So, the number of moles of #H^+# in our acid is:
#"0.0316 mol/L" xx "0.200 L" = "0.00632 mol"#
This means that the volume of #"12.2 M"# (or #"12.2 mol/L"#) #HCl# needs to have #"0.00632 mol"# of #H^+# ions.
#HCl# is a strong acid, so #1# mole of #HCl# is virtually the same thing as #1# mole of #H^+# ions.
To have #0.00632# moles of #HCl#, the volume must be:
#"0.00632 mol" / "12.2 mol/L" = "0.000518 L"#
Since #HCl# is a strong acid, the same amount must be needed to have #0.00632# moles of #H^+# ions—#"0.000518 L"#.
Finally, #"0.000518 L"# in #"mL"# is #"0.518 mL"#.
