Question

Polyprotic acids have more than one proton to donate to water, therefore have more than one equilibrium constant for proton donation. For phosphoric acid, there is a three-step equilibrium: H, PO +H,0 5 H2PO4 + H30* H,PO," +H,0 5 HPO - +H30* HPO 2- +,0 5 PO.- +,0* Kai = 7.11 x 10-3 Ka2 = 6.32 x 10-8 Ka; = 4.5 x 10-13 For all conjugate acid base pairs: K, XKK where K is for the reaction of the conjugate base in water to produce OH-ions. This relationship allows us to calculate either the Kb of the anion in a basic salt from the K, of its conjugate acid or the K, of the cation in an acidic salt from the Ko of its conjugate base,

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Answer 1

MARCH S M T W 1 22 10 W0AW Wednesday 09.00 2._cuilibrius Step_important for determin -Naz Pou sol is or *+7-11X1oºx # -1.42x10°-o 2 = 0.034 0900 10.00 3- = POU+150 Hpoů + oHº 10.00 50, P12 -log_[] = -log[0.034) = 1:47 1100 1.00 12.00 H₂ POy is tri baric aud, Therefore has three _dussociation constant however_1_second and third dissociation constant is very small therefore negligible. Therefore we must consider the concentration obh ion_only from dissociation of_15+ Step in deriving pH of the sot". - 12.00 pH of 0. 2M Na, PO u 501 Naz pou & Na + pou I PO? _.POMPEHO = H Pozł +046 Initially 0.2 o o (0.2-x) x * 1.00 0.00 0.00 H_ a 02.00 _initials_02 _ 0 0 Kreupon? ][ pop -I PO43-I _ x (0:2-8) 2 03.00 14 - 75 Pou + 160 Hz pou +150 Rai = 7.11710 (0-2-2) Ka.[poy][130]; 7.11810 - [hy P04- x? 7-11x40-9.fassume txx<02) 70-2-x)^*'" en T TH TO H ono o Now, Kyo Ka 1810 . 0.02 4.5+ 10-13 7.11X10 4.5x10-7320-02 04.00 3 xk 30, 05.00 06.00 PHIL = 0.02_ 0.913) 30, 36 = 0.054 - 50o -0-054-72043-log(ol) = . __ Py = 14-POH = 14- 1.27 =12.83 4. NaF Baric kory Neutral Na2CO₃- Acidic KF Basic CCOONa → Basic Nal Neutred Naou Neutral NHyce Acidic NaNO₂ Neutral Nq5o Basic - Acidic Natco₂ Basic Na,o- Basic Coop24.1JAJOS Smember il C COONA T dr, az Baclyt