A 25.0 ml sample of 0.150 M nitrous acid is titrated with a0.150 M NaOH solution. What is the pH at the equivalence point? theKa of Nitrous Acid is 4.50 x 10-4.

3 Answers
SCooke
Apr 16, 2018

See the answer by @Andrea. https://socratic.org/users/andrea-b-3

Andrea B.
Apr 16, 2018

pH= 8.11

Explanation:

When the titration occurrs you have used 25,0 mL of NaOH so you in total have 50.0 mL of #NaNO_2# that (because the volume is double) will be 0.075 M
pH of equivalent point is the pH of this solution of #NaNO_2# 0.075 M
It is a salt with a basic hydrolysis
#[OH^(-)]= sqrt (K_w/K_a xx C_s)=sqrt((1 xx 10^(-14))/(4.5 xx 10^(-4)) xx 7.5 xx 10^(-2) )= sqrt(1,67 xx 10^(-12))= 1.29 xx 10^-6#
pOH = 5.89
pH= 8.11

Truong-Son N.
Apr 17, 2018

You should treat this problem with care... most students would #(i)# use the Henderson-Hasselbalch equation and use #"HNO"_2# as the acid and #"NaOH"# as the base, #(ii)# use the #K_a# and thus assume #"HNO"_2# is dissociating in water, or #(iii)# forget about dilution and use #"0.150 M" = ["NO"_2^(-)]_i#. That doesn't make sense.

The #"pH"# is #8.11#.

By neutralizing to the equivalence point, only #"NO"_2^-# remains. That comes from

#"0.150 mol HNO"_2/"L" xx "0.025 L" = "0.00375 mols NO"_2^(-)# formed

upon exact neutralization of #"HNO"_2# with #"NaOH"#. This is now in a volume of #"25.0 mL" + "25.0 mL" = "50.0 mL" = "0.050 L"#.

Therefore, we get a diluted concentration of

#["NO"_2^-]_i = "0.00375 mols"/"0.050 L" = "0.075 M NO"_2^-#,

as expected since the volume was doubled (so the concentration should halve).

Now, this nitrite will associate in an equilibrium within water:

#"NO"_2^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HNO"_2(aq) + "OH"^(-)(aq)#

#"I"" "0.075" "" "" "" "-" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "" "-" "" "+x" "" "+x#
#"E"" "0.075-x" "" "-" "" "x" "" "" "" "x#

Here we have a BASE, and so we must use the #K_b#:

#K_aK_b = K_w#

#=> K_w/K_a = K_b = 10^(-14)/(4.50 xx 10^(-4)) = 2.22 xx 10^(-11)#

It is apparent that this #K_b# is very small, so the small #x# approximation works here very well. This is true usually when #K# is on the order of #10^(-5)# or less.

Write the mass action expression:

#K_b = 2.22 xx 10^(-11) = x^2/(0.075 - x) ~~ x^2/0.075#

Hence,

#x -= ["OH"^(-)] = sqrt(0.075K_b)#

#= sqrt(0.075 cdot 2.22 xx 10^(-11))#

#= 1.29 xx 10^(-6) "M OH"^-#

Therefore,

#"pOH" = -log["OH"^(-)] = 5.89#

#color(blue)("pH") = 14 - "pOH" = color(blue)(8.11)#

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