Answer :
Answer:
The volume of KOH needed for complete neutralization is 2.143 ml
Explanation:
Complete neutralization reaction is given below as;
H₂SO₄ + 2KOH ---------> K₂SO₄ + 2H₂O
Acid to base ratio, na : nb = 1 : 2
[tex]\frac{C_AV_A}{C_BV_B} =\frac{na}{nb}[/tex]
where;
[tex]C_A[/tex] is the concentration of the acid ( H₂SO₄) = 0.25 M
[tex]C_B[/tex] is the concentration of the base (KOH) = 0.35 M
[tex]V_A[/tex] is the volume of the acid = 15 ml
[tex]V_B[/tex] is the volume of the base = ?
[tex]\frac{0.25*15}{0.35*V_B} =\frac{1}{2}\\\\0.35*V_B =2(0.25*15)\\\\0.35*V_B = 7.5\\\\V_B = \frac{7.5}{3.5} = 2.143 ml[/tex]
The volume of KOH needed for complete neutralization is 2.143 ml