What is the molarity of an naoh solution if 3.47 ml is titrated by 11.1 ml of 0.0904 m hno3?

V ( NaOH ) = 3.47 mL in liters =  3.47 / 1000 => 0.00347 L

M ( NaOH) = ?

V ( HNO₃) = 11.1 / 1000 => 0.0111 L

M ( HNO₃) = 0.0904 M

number of moles HNO₃ :

n = M x V = 0.0904 x 0.0111 => 0.00100344 moles of HNO₃

mole ratio :

NaOH + HNO₃ = H₂O + NaNO₃

1 mole NaOH 1 mole HNO₃
(moles NaOH )  0.00100344 moles HNO₃

moles NaOH = 0.00100344 x 1 / 1

moles NaOH = 0.00100344 / 1

= 0.00100344 moles of NaOH

M ( NaOH ) = n / V

M ( NaOH ) = 0.00100344 / 0.00347

= 0.289 M

hope this helps!

For titration, the number of equivalence of the solutions used should be equal. For the given, that is,
                                Ma x Va = Mb x Vb
Substituting the known values,
               (11.1 mL) x (0.0904 M HNO3) = (3.47 mL) x Mb
The value of Mb is approximately equal to 0.289 M NaOH.

If a hydrogen atom is neutral, it must also have  one electron. an oxygen atom (o) has an atomic number of eight (8). this means a neutral oxygen atom has eight protons and eight electrons. the element actium (ac) has an atomic number of 89, so it has 89 protons and 89 electrons in a neutral atom

Sorry i dont understand the question can you rephase it?