What volume (mL) of 0.386 M potassium hydroxide solution would just neutralize?

Question

What volume (mL) of 0.386 M potassium hydroxide solution would just neutralize 64.9 ml of 0.379 M H2SO4 solution?

Answer 1

to neutralize NaOH you must use two times less H2SO4since the acid is divalent and KOH monovalent.
H2SO4+2 KOH--> K2SO4 + 2H2O

So .386M of KOH neutralizes 0.193M of H2SO4

here you have a solution of .379

So, the ratio volume of KOH /volume H2SO4 is 0.379/0.193=1.964

and as you have 64.9 mL
you must use 64.9*1.964=127.4mL KOH0.386M

Answer 2

The rection is two Moles of KOH reacts with one mole H2SO4, for the reason that each KOH has one OH to make a contribution, and each mole of H2SO4 has 2 moles H to make a contribution. 2KOH + H2SO4 --->K2SO4 + 2H20 reaction fee is two:a million, please be conscious this on the top of the calculation! N1V1 = N2V2 (0.348 M KOH) (X Vol mL) = (0.219 M H2SO4) (119.4 mL) X Vol mL KOH = (0.219 M H2SO4) (119.4 mL)/(0.348 M KOH) X Vol mL KOH = seventy 5.14 mL x 2 = a hundred and fifty.3 mL explanation why we ought to multiply via 2 in the top is that it takes two times as plenty KOH to reaction with 0.5 the quantity of H2SO4 via volume.

Answer 3

2x64.9x0.379/0.386 = 127.45 mL