Chemistry, asked by config9378, 9 months ago

The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (HOOCCOOH) and must be removed before the stems are used to make rhubarb pie. If pKa1 = 1.23 and pKa2 = 4.19, what is the pH of a 0.0200 M solution of oxalic acid?

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Answered by Pampa21
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Answer:Concentrati on of H2C20.0247 M ICE table constructedis as follows HHC(aq) + H,o'(aq) I(M)00247 C(M) x E (M) (00247-x) 1st acidionizati on constant, Kal =_THAO4 10-pr-10-123-5.9810-2- (00247-x x0.0187 From the equilibrium table, [H3ợJ-x:0 0187M [HC204-]-x=0.0187M ICE table constructed for the2 onization of H2C0 is as follows HCO,-(aq) + H20(1) 992-(aq) + H,0+(aq) 0,042-(aq) H,0+ (aq I(M): 0.0187 C(M): -x E (M): (0.0187-x) 0.0187 0.0187+x 2"d acidionization constant, K2 10-p":10-4."-6 4x10% (0.0187 +x)(x) 0.0187- x x= 6.4×10-5 From the equilibrium table. [H(0.0187+x)M =(0.0187+6.4×10-5 0.0187 M pH-l0g[H,0log (0.0187) 1.73

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