b) 100 ml that 0.1 M equipment of ammonium chloride come which 100 ml of0.1 M solution of salt hydroxide.

You are watching: Calculate the ph of a 1 mnh4cl solution.

Assume that all volumes space additive. Kb (ammonia) = 1.8 x 10-5

For a):

Ammonium chloride, #NH_4Cl# disappear in solution to form ammonium ions #NH_4^(+)# i m sorry act as a weak acid by protonating water to kind ammonia, #NH_3(aq)# and hydronium ion #H_3O^(+)(aq)#:

#NH_4^(+)(aq)+H_2O(l) -> NH_3(aq)+H_3O^(+)(aq)#

As we know the #K_b# because that ammonia, us can discover the #K_a# for the ammonium ion. Because that a offered acid/base pair:

#K_a time K_b=1.0 times 10^-14# assuming conventional conditions.

So, #K_a(NH_4^(+))=(1.0 times 10^-14)/(1.8 time 10^-5)=5.56 time 10^-10#

Plug in the concentration and also the #K_a# value right into the expression:

#K_a=( times (NH_3>)/()#

#5.56 times 10^-10~~( times )/(<0.1>)#

#5.56 time 10^-11=^2#

(as we deserve to assume that one molecule hydronium must type for every one of ammonia the forms. Also, #K_a# is small, for this reason #x ≪ 0.1#.)

#=7.45 times 10^-6#

#pH=-log#

#pH=-log(7.45 time 10^-6)#

#pH approx 5.13#

For b):

(i) identify the varieties present after mixing.

The equation because that the reaction is

#color(white)(mmmmm)"OH"^"-" + "NH"_4^"+" -> "NH"_3 + "H"_2"O"##"I/mol": color(white)(mll)0.010 color(white)(mll)0.010color(white)(mol/L)0##"C/mol": color(white)(m)"-0.010"color(white)(ml)"-0.010"color(white)(m)"+0.010"##"E/mol": color(white)(mll)0color(white)(mmmm)0color(white)(mmml)0.010#

#"Moles the OH"^"-" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol"#

#"Moles that NH"_4^"+" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol"#

So, us will have 200 mL of one aqueous equipment containing 0.010 mol the ammonia, and the pH have to be higher than 7.

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(ii) calculate the pH the the solution

#<"NH"_3> = "0.010 mol"/"0.200 L" = "0.050 mol/L"#

The chemistry equation for the equilibrium is

#"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"#

Let"s re-write this as

#"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"#

We can use an ice table to perform the calculation.

#color(white)(mmmmmmmll)"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"##"I/mol·L"^"-1":color(white)(mll)0.050color(white)(mmmmmll)0color(white)(mmm)0##"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmll)"+"xcolor(white)(mll)"+"x##"E/mol·L"^"-1":color(white)(m)"0.050-"xcolor(white)(mmmmm)xcolor(white)(mmm)x#