12.5ml of 2.0m NaOH neutralizes 10ml og HCL, what is the acid concentration

2.5M

Explanation:

We begin by writing a balanced equation for the reaction. This is shown below:

HCl + NaOH —> NaCl + H2O

From the equation above we obtained the following:

nA (mole of the acid) = 1

nB (mole of the base) = 1

Data obtained from the question include:

Vb (volume of the base) = 12.5mL

Mb (Molarity of the base) = 2M

Va (volume of the acid) = 10mL

Ma (Molarity of the acid) =?

The molarity of the acid can be calculated as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 2 x 12.5 = 1

Cross multiply to express in linear form as shown below:

Ma x 10 = 2 x 12.5

Divide both side by 10

Ma = (2 x 12.5) /10

Ma = 2.5M

The molarity of the acid is 2.5M

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer