Question

An aliquot (28.7 mL) of a KOH solution required 31.3 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was in the original sample? Question 4 options: A) 0.173 B) 7.28 C) 1.64 D) 0.207 E) 0.414

Answer 1

= 28.7 m2 IL Sol=> An aliquot (28.7 m2) of a kot solution required 31.3 mL of 0.118 m Hel for neutralization, what mass cym) of Kol was in the original sumple 2. => correct option is CD i.e.o.207 Calculation - Given! - volume of Koh solution - 28.7 20 x -0.0287 | 1000 mL volume of Hel solution = 31.3 mL = 31.3 mLt 1 L 1000 mot concentretion (molimity) of Hcl = 0.118 M = 0.118 mol/L Baleineed equation for this reeeltion :- Kohcoat Hel kel Cag) = If Calculate number of moles of Hel first.- molarity = number of moles iema n volume of solution in Rearrange this formula to cakukute n. na m® V = 0.118 mole/Z x 0.0313 Ł -0.0313 L can of H2O (d) v n = 0.0036934 mole = number of moles of HCl = 0.0036934 mole

=> 2) converts moles of Hel into moles of koh:- According to baluneed equation 1 mole Hel neutralized requires 1 mole koH. 1 mole koh -0.0036934 mole Hal x 1 molettel ·0036934 mole koH. =) number of moles of koH = 0.0036934 mole. => 3) converts moles of Koh into gm of koh:- grams = moles y molar muss Co. Molur mass of Koh = 56.11 9m/mole) grams = 0.0036934 moté x 56.41 gm/ møte grams = 0. 207 you =) mass of Kol was in original sumple = 0.207 gm =) Henes correct option is CD