Determine the initial pH, the pH at the midpoint (halfway to the equivalence point), and the pH at the equivalence point if 25.00mL of 0.1335 M NH3 (NH4OH) is titrated with 0.2350M HCl? Kb = 1.75*10(-5)

Question

Luna Chen · Accepted Answer

(a). 11.2

(b). 9.24

(c). 5.16

#NH_(3(aq))+H_2O_((l))rarrNH_(4(aq))^(+)+OH_((aq))^-#

Initial moles #NH_3# = 0.1335

Initial moles #OH^-# = 0

If #x# moles of #NH_3# are used up then:

Equlibrium moles #NH_3 =( 0.1335 - x)#

Equilibrium moles #OH^-# = #x#

#K_b=([NH_4^(+)][OH^-])/([NH_3])=1.75xx10^(-5)#

Since #x# is very small we assume #(0.1335=x)rarr0.1355#

So: #([OH^-]^(2))/(0.1335)=1.75xx10^(-5)#

From which #[OH^-] = 1.53xx10^(-3)M#

#pOH=2.815#

#pH=pK_w-pOH#

pH = 14 - 2.815 = 11.2

At the mid - point of the titration #[NH_4^+]=NH_3]#

So #K_b=[OH^-]=1.75xx10^(-5)#

#[H^+]=(K_w)/([OH^-])=(10^(-14))/(1.75xx10^(-5))=5.714xx10^(-10)#

#pH=-log(5.714xx10^(-10))#

pH is 9.245 in this case.

We must now obtain the entire volume.

#NH_3+HClrarrNH_4Cl+H_2O#

moles #NH_3 = (0.1335xx25)/(1000)=3.34xx10^(-3)#

So moles #HCl=3.34xx10^(-3)#

#c=n/v#

#v=n/c = (3.34xx10^(-3))/(0.235)L=14.2ml#

So total volume = #25 +14.2 = 39.2ml#

The final pH will be slightly acidic due to salt hydrolysis of #NH_4^+#:

#NH_4^(+)rightleftharpoonsNH_3+H^+#

no. moles #NH_4^(+)=3.34xx10^(-3)#

So #[NH_4^(+)]=(3.34xx10^(-3))/(39.2xx10^(-3))=0.0852M#

#K_a=([H^+]^2)/(0.0852)#

We know that #K_a=K_w/K_b=(10^(-14))/(1.75xx10^(-5))=5.7xx10^(-10)#

So #([H^+]^2)/(0.0852)=5.7xx10^(-10)#

#[H^+]=6.968xx10^(-6)#

#pH=-log(6.968xx10^(-6))#

#pH=5.16#

Noah Aldrich · Answer

undefined Initial pH: 11.47
pH at midpoint: 9.61
pH at equivalence point: 4.11