A 26.8 mL sample of a solution of RbOH is neutralized by 15.71 mL of a 1.918 M solution of HBr. What is the molarity of the RbOH solution?
RbOH + HBr ==> RbBr + H2O
mols HBr = M x L = ?
mols RbOH = mols HBr
M RbOH = mols HBr/L HBr = ?
Also, because the reaction is a 1:1 rxn ratio, it is useful to apply (Molarity x Volume)Acid = (Molarity x Volume)Base => (1.918M)(15.71ml) = M(base) x (26.8ml). Solve for Molarity of base (RbOH). The volumes can be left in ml as both sides of the equation contain mls, they will cancel and only Molarity remains.
To determine the molarity of the RbOH solution, you can use the equation:
Molarity (RbOH) x Volume (RbOH) = Molarity (HBr) x Volume (HBr)
In this case, the volume of the RbOH solution is 26.8 mL, and the volume of the HBr solution is 15.71 mL.
Let's substitute the values into the equation:
Molarity (RbOH) x 26.8 mL = 1.918 M x 15.71 mL
Now we can solve for the molarity of the RbOH solution:
Molarity (RbOH) = (1.918 M x 15.71 mL) / 26.8 mL
Calculating this equation will give you the molarity of the RbOH solution.
To solve this problem, we will use the concept of stoichiometry. Stoichiometry is the study of quantitive relationships between reactants and products in a chemical reaction.
First, let's write a balanced equation for the neutralization reaction between RbOH and HBr:
RbOH + HBr → RbBr + H2O
Next, we need to determine the number of moles of HBr used in the neutralization reaction. To do this, we can use the equation:
moles of solute (HBr) = molarity × volume (in liters)
Given:
Molarity of HBr solution (M1) = 1.918 M
Volume of HBr solution (V1) = 15.71 mL = 15.71/1000 = 0.01571 L
moles of HBr = 1.918 M × 0.01571 L
Next, we need to determine the number of moles of RbOH that reacted with the HBr. According to the balanced equation, the stoichiometric ratio between HBr and RbOH is 1:1. Therefore, the moles of RbOH used in the reaction will be the same as the moles of HBr.
moles of RbOH = moles of HBr
Now, we can calculate the molarity of the RbOH solution.
Given:
Volume of RbOH solution (V2) = 26.8 mL = 26.8/1000 = 0.0268 L
Molarity of RbOH solution (M2) = ?
moles of RbOH = moles of HBr = 1.918 M × 0.01571 L
Molarity of RbOH solution (M2) = moles of RbOH / Volume of RbOH solution (V2)
Substituting the values, we get:
M2 = (1.918 M × 0.01571 L) / 0.0268 L
Now, perform the calculation:
M2 = 1.129 M
Therefore, the molarity of the RbOH solution is 1.129 M.