Question

Calculate the pH change that results when 12 mL of 5.9 M NaOH is added to 753 mL of each the following solutions. (See the appendix.)

A) 0.10 M NH₄Cl


B) 0.10 M NH₃

Acid Name	ConjugateAcid	Ka	pKa	Conjugate Base	Base Name
perchloric acid	HClO4	>>1	< 0	ClO41-	perchlorate ion
hydrohalic acid	HX (X=I,Br,Cl)	>>1	< 0	X1-	halide ion
sulfuric acid	H2SO4	>>1	< 0	HSO41-	hydrogen sulfate ion
nitric acid	HNO3	>>1	< 0	NO31-	nitrate ion
hydronium ion	H3O1+	1.0	0.00	H2O	water
iodic acid	HIO3	0.17	0.77	IO31-	iodate ion
oxalic acid	H2C2O4	5.9 x 10-2	1.23	HC2O41-	hydrogen oxalate ion
sulfurous acid	H2SO3	1.5 x 10-2	1.82	HSO31-	hydrogen sulfite ion
hydrogen sulfate ion	HSO41-	1.2 x 10-2	1.92	SO42-	sulfate ion
phosphoric acid	H3PO4	7.5 x 10-3	2.12	H2PO41-	dihydrogen phosphate ion
hydrofluoric acid	HF	7.2 x 10-4	3.14	F1-	fluoride ion
nitrous acid	HNO2	4.0 x 10-4	3.40	NO21-	nitrite ion
lactic acid	HC3H5O3	6.4 x 10-5	3.85	C3H5O31-	lactate ion
formic acid	HCHO2	1.8 x 10-4	3.74	CHO21-	formate ion
hydrogen oxalate ion	HC2O41-	6.4 x 10-5	4.19	C2O42-	oxalate ion
hydrazoic acid	HN3	1.9 x 10-5	4.72	N31-	azide ion
acetic acid	HC2H3O2	1.8 x 10-5	4.74	C2H3O21-	acetate ion
carbonic acid	H2CO3	4.3 x 10-7	6.37	HCO31-	hydrogen carbonate ion
hydrogen sulfite ion	HSO31-	1.0 x 10-7	7.00	SO32-	sulfite ion
hydrosulfuric acid	H2S	1.0 x 10-7	7.00	HS1-	hydrogen sulfide ion
dihydrogen phosphate ion	H2PO41-	6.2 x 10-8	7.21	HPO42-	hydrogen phosphate ion
hypochlorous acid	HClO	3.5 x 10-8	7.46	ClO1-	hypochlorite ion
ammonium ion	NH41+	5.6 x 10-10	9.25	NH3	ammonia
hydrocyanic acid	HCN	4.0 x 10-10	9.40	CN1-	cyanide ion
hydrogen carbonate ion	HCO31-	4.7 x 10-11	10.33	CO32-	carbonate ion
hydrogen phosphate ion	HPO42-	4.8 x 10-13	12.32	PO43-	phosphate ion
hydrogen sulfide ion	HS1-	1.3 x 10-13	12.89	S2-	sulfide ion
water	H2O	1.0 x 10-14	14.00	OH1-	hydroxide ion
ammonia	NH3	<<10-14		NH21-	amide ion
hydroxide ion	OH1-	<<10-14		O2-	oxide ion