Respuesta :
If 25.0ml neutralizes 35.0 ml of a 0.200M HCl solution, the molarity of KOH solution is 0.28M
The reaction between KOH solution and HCl is written below.
[tex]HCl + KOH \rightarrow KCl + H_2O[/tex]
Volume of KOH, [tex]V_B[/tex] = 25.0 ml
Volume of HCl, [tex]V_A=35.0 ml[/tex]
Molarity of HCl, [tex]C_A=0.200 M[/tex]
Molarity of KOH, [tex]C_B=?[/tex]
Number of moles of HCl, [tex]n_A=1[/tex]
Number of moles of KOH, [tex]n_B=1[/tex]
The mathematical equation for neutralization reaction is:
[tex]\frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B}[/tex]
Substitute [tex]C_A=0.200 M[/tex], [tex]n_A=1[/tex], [tex]n_B=1[/tex], [tex]V_B[/tex] = 25.0 ml, and [tex]V_A=35.0 ml[/tex] into the equation above in order to solve for [tex]C_B[/tex]
[tex]\frac{0.2 \times 35}{C_B \times 25}=\frac{1}{1} \\\\25C_B=7\\\\C_B=\frac{7}{25} \\\\C_B=0.28M[/tex]
Therefore, the molarity of KOH solution is 0.28M
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