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Sahay Sir > Question Answers > N Awasthi Physical Chemistry > Chapter 6 - Ionic Equilibrium (Level 1 and Level 2) > 50 mL of 0.05 M Na2CO3 is titrated against 0.1 M HCl. On adding 40 mL of HCl, pH of the solution will be [Given for H2CO3, pKa1 =6.35; pKa2 ​= 10.33; log3=0.477, log2=0.30]

50 mL of 0.05 M Na2CO3 is titrated against 0.1 M HCl. On adding 40 mL of HCl, pH of the solution will be [Given for H2CO3, pKa1 =6.35; pKa2 ​= 10.33; log3=0.477, log2=0.30]

50 mL of 0.05 M Na2CO3 is titrated against 0.1 M HCl. On adding 40 mL of HCl log2=0.30] pH of the solution will be [Given for H2CO3 pKa1 =6.35; pKa2 ​= 10.33; log3=0.477