Moles of HNO₂ (0.16) (50) (10³) · 5x10³ moles Moley of hot (0.15) (25) (16¹) = 3.75 x 10³ mokes 12. 50.00 mL of 0.10 M H

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Moles Of Hno 0 16 50 10 5x10 Moles Moley Of Hot 0 15 25 16 3 75 X 10 Mokes 12 50 00 Ml Of 0 10 M H 1 (35.96 KiB) Viewed 43 times

Moles of HNO₂ (0.16) (50) (10³) · 5x10³ moles Moley of hot (0.15) (25) (16¹) = 3.75 x 10³ mokes 12. 50.00 mL of 0.10 M HNO₂ (nitrous acid, K₂ = 4.5 x 10-4) is titrated with a 0.15 M KOH solution. After 25.00 mL of the KOH solution is added, the pH of the solution will be 1.29-15 HINO₂ + LDA KNO₂+ +1₂0 [HD₂] = PH=pha + log [₂] [UNDE] I Sand 3.45 75 0.06x16³M a 2.17 Opel = C-3.75 ml -3.756² m² - +3.75 +10 E 1.25 3.35 = [log (4.5 10*)] + [10g/005163) PH= 3.84 (004²3) Ô 3.05 1 C 2.41 [KAR] = 1.75² d. 3.82 25 e. 7.00 13. What is the volume of KOH needed to reach the equivalent point in the previous question? 50.1 = x(0.15) a. 50.0 mL At equivalence point b. 75.0 mL x= volume of KOH x=33.33md C33.3 mL 83.3 mL d. Milimoles of [H00₂]: milimble of KOH e. 47.0 mL 14. In class you learned that during the titration of a weak acid with a strong base the salt that is produced in the reaction will react with water. Knowing this, could you calculate the pH at the equivalent point related to the previous question? a. 8.06 b. 8.33 C. 5.94 d. 11.71 e. 9.02 b. مرقند ۵۸ -