Question

A 0.3125 g sample of a primary Na2CO3 standard was treated with 40.00 mL of dilute perchloric acid. The solution was boiled to remove Co 2, and thereafter the excess HClO 4 was titrated by recoiling with 10.12 mL of dilute NaOH. In a separate experiment, it was established that 27.43 mL of HClO4 neutralized the NaOH in a 25.00 mL portion. Calculate the molarities of HClO4 and NaOH

Answer 1

25 0.25 ml NaOH = 27.43 m2 HU04 0.31259 Na2CO3 + 10:12 ml NaOH = 40 ml H6104 op 0.31259 NQ,C03 + 27:43 x 10.12 m2 HC10A = 40 ml HUOA on 0.3125 g Na2CO3 = (40 - 27:43.410.127 ml HC104 = 28.90 mL HUDA Now, Na2CO3 + HU104 = NACI04 + 602 + H2O 1 mol i mol : ; 0.3125 9 Na2CO3 = 0.3125 moli Na2CO3 [Molar mas's 105.9888 of Na2CO3 = 105.98887 =10.3125 g mol. HUOA =lios 105.9888 .: Molarity of HU0A = 1013.13588 * 2009.) = 0.102(M). VNQOH * CNaOH = VHOAX CH4104 25 (ml)x e Nach = 27:43 (m) x 0.102(M) op (NaOH = 27.43 x 0-102 (M) = 0.112(M). i Molarity of HU104 = 0.102CM) Zames and ggº 79 NaOH = 0.112 CM) Y (Any) - 25