As the feedback and solution both indicate, the difference in calculate #"pH"# once solving through the quadratic equation and also the presumption that #x# is little enough different by only #0.01# #"pH"# units, so making use of the #x#-is-small presumption is recommended.

#K_a = 1.8 * 10^(–4)# in ~ #25^

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Formic mountain is a **weak acid**, therefore you know that the will just partially ionize in aqueous systems to create formate anions and also hydronium cations.

#"HCOOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCOO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#

Notice that every #1# **mole** that formic acid **that ionizes** produces #1# **mole** of formate anions and #1# **mole** of hydronium cations.

Now, let"s say the #x# #"M"# is the concentration the formic acid **that ionizes**. Because formic mountain is a weak acid, you understand for a truth that

#x quad "M"

In various other words, only a fraction of the early concentration that the acid will certainly actually ionize. Follow to the balanced chemical equation, the solution will save

#<"HCOO"^(-)> = <"H"_ 3"O"^(+)> = x quad "M"#

When #x# #"M"# that formic mountain **ionizes**, you acquire #x# #"M"# the formate anions and also #x# #"M"# that hydronium cations.

So **at equilibrium**, you have the right to express the concentration the the formate anions and the concentration the the hydronium cations in regards to the concentration that formic mountain **that ionizes**. At equilibrium, the resulting solution will also contain

#<"HCOOH"> = (0.0700 - x) quad "M"#

When #x# #"M"# the formic acid **ionizes**, the early concentration the the acid **decreases** through #x# #"M"#.

See more: Which One Of The Following Compounds Contains Ionic Bonds? ?

By definition, the **acid dissociation constant** that explains the ionization equilibrium of formic acid is equal to

#K_a = (<"HCOO"^(-)> * <"H"_ 3"O"^(+)>)/(<"HCOOH">)#

In your case, this will be same to

#K_a = (x * x)/(0.0700 - x)#

which is

#1.8 * 10^(-4) = x^2/(0.0700 - x)#

Now, you understand that making use of the approximation

#0.0700 - x ~~ 0.0700#

is encourage here, for this reason rewrite the expression of the mountain dissociation consistent as

# 1.8 * 10^(-4) = x^2/0.0700#

This will obtain you

#x = sqrt(0.0700 * 1.8 * 10^(-4)) = 0.003550#

Since #x# to represent the **equilibrium concentration** that hydronium cations, you have the right to say the the solution will have

#<"H"- 3"O"^(+)> = "0.003550 M"#

As girlfriend know, the #"pH"# that the equipment is offered by

#"pH" = - log in (<"H"_3"O"^(+)>)#

Plug in your worth to find

#"pH" = - log(0.003550) = color(darkgreen)(ul(color(black)(2.450)))#

The price is rounded to 3 **decimal places** since you have three **sig figs** for the early concentration the formic acid.