Question

Question #672a6

Answer 1

`"pH" = 4.56`

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You have to look up the `K_b` of this compound's conjugate base, which is about `1.3 xx 10^(-6)`. Alternatively, the `"pKa"` of hydrazinium is listed on the hydrazine Wikipedia as `8.10`. However, you should use the `K_a` listed in your textbook to be more accurate.

I'll use the `K_b` since it is more educational. Recall that:

`K_w = K_aK_b`.

Therefore, you can get the `K_a` of the acid, which is what you are actually placing into water. Remember to always use `K_a` for acids and `K_b` for bases.

`K_a = 10^(-14)/(1.3 xx 10^(-6)) = 7.69 xx 10^(-9)`

The reaction itself can be denoted in general:

`"BH"^(+)(aq) " "" "+" "" " "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "B"(aq)`

`"I"" ""0.10 M"" "" "" "" "" "" "-" "" "" ""0 M"" "" "" "" ""0 M"`
`"C"" "-x" "" "" "" "" "" "" "-" "" "+x" "" "" "" "+x`
`"E"" "(0.10 - x)"M"" "" "" "" "-" "" ""x M"" "" "" "" ""x M"`

And its `K_a` can be written with a mass action expression:

`K_a = x^2/(0.10 - x)`

However, since `K_a "<<" 10^(-5)`, it's a great time to make the small `x` approximation. That is, if you're tired of using the quadratic formula.

`K_a ~~ x^2/(0.10)`

Therefore, we can solve for the equilibrium concentration of `"H"^(+)` easily:

`["H"_3"O"^(+)] = x ~~ sqrt(0.10K_a)`

`= 2.77 xx 10^(-5) "M"`

Hence, the `"pH"` is:

`color(blue)("pH") = -log["H"_3"O"^(+)]`

`= color(blue)(4.56)`

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And another way to check whether the small `x` approximation is appropriate is to see if the percent dissociation is less than `5%`.

`%"dissoc". = x/(["soln"])`

`= (2.77 xx 10^(-5) "M")/("0.10 M") xx 100% = 0.028% "<<" 5%`

So this approximation works extremely well. You would get pretty much the same answer if you chose to solve the quadratic equation in full.