Question

50.0 mL of 0.090 M nitrous acid (HNO₂, K_a = 7.1 x 10^-4), is titrated with 0.100 M NaOH, requiring 45.0 mL of strong base to reach the equivalence point.

(a) What will be the pH after 35.0 mL of NaOH have been added?

(b) What will be the pH at the equivalence point?

(c) What will be the pH after 60.0 mL of NaOH have been added?

Answer 1

Titration of Weach a aud with • o neutralisation reaction takes place between nitrous and (HMO2) and strong base odium hy dronide (MaoH). It the neutralisation is not Complete more specifically, if the base is not completely neutralised , we have a buffer solution that will contain nitrous and and its confurare base (Moz) in. groles of Jino . molarity- melye con 7 50.00mL of o.ogom solution % чор , с хvно, in noles 0.09 moll of LL 0.0045 mole Naast in 35.0mL of a.loon Solution hrach = cxv. o.1 oomole. 1000 L so 5 mol 1000L The nentralisation reaction, ICE table at No2 Cep t auton Initial (moly 8.00 45 0.0035 change moles – 0.0035 -0.0035 Equilibrium (more orool o Total volume during No2 (ay + Holan +000035 0.0035 neutralisation 50.00 +35 mL con un trations of each speies; 0.oo Imole 0.0035 more [ HNO2] = 0.1852 ; [nop] = 0 Henderson. Hasselbalch equations, Lou 85m2 20.0854 mo,7 - 0.0851 According to Henderson- pH =pha tlog [Base >> pu- logika) top (nou) sp -log(+37104) thing to the >> pho-logi Base) tudy CN) 10.00 0-085)

- pH = 3.1487 + 0.544 - pH = 3.6927 - pH of solution after addition of 3 of ma oH = 3.6927 Nach in 45.0m2 of o.100 m Solution. haolt = cхr - 01 Do inole At equivalence, i 1L 1000L = 0.0045 male At equivalence point mous of acid must be equal to mous of base added The neutralisation neartion, ICE tible Hng (an tot cam Mon(at) + H₂O bitial role 0.0045 0.0045 Change mole) -0.0045 -0.0045 +0.0045 Equilibrio trole) 0 0 Total volume of solution; = 50+450m2 i Molar concentration, = 95m2 [10,-) - 0.0045 mode ( = 0.0951 0.095 L 0.0045 Noz don is the = 0.04736M Noz son is the conturate base of weak acid Amo, It dissociates in asnang solution as follows ICE table ng Car & H₂O(l) HNO, (ar) totam Initialing 0.04736 - -n Changen - ta ta Equilibrium (m) (0.04736x) - - In be the amount of change, By definition equilibrium constant expression: woo Etnog] [aut-] S kaxkg = new [ Mor - >) 1.408 X107. xx* 7.1 x64 (0.04736-x = kg - 10-14 2) ko = 1.408 x 10" since no is very sman; (6.04736-n) - 0.0473x