A 35.6 mL solution of Sr(OH)2 is titrated with 0.549 L of 0.0445 M HBr. Determine the concentration of strontium hydroxide.

The answer to your question is  0.34 M  

Explanation:

Data

[Sr(OH)₂] = ?

Volume of Sr(OH)₂ = 35.6 ml or 0.0356 l

[HBr] = 0.0445 M

Volume of HBr = 0.549 l

Balanced chemical reaction

                 2HBr  +  Sr(OH)₂  ⇒  SrBr₂   +   2H₂O

Process

1.- Calculate the moles of HBr

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = 0.0445 x 0.549

          = 0.0244

2.- Calculate the moles of Sr(OH)₂ using the coefficients of the balanced reaction

                    2 moles of HBr 1 mol of Sr(OH)₂

                    0.0244 moles    x

                                   x = (0.0244 x 1) / 2

                                   x = 0.0122 moles of Sr(OH)₂

3.- Calculate the concentration of Sr(OH)₂

Molarity = 0.0122/ 0.0356

-Simplification

Molarity = 0.34                      

idk what is it? is this a question that you know the answer or u dont