11.4mL of 0.7 M NaOH was used to titrate 5.0mL hydrochloric acid. What is the molarity of the acid?
a) 0.3
b) 1.6
c) 3.9
d) 5.0
e) 8.1
Can someone answer this question and explain how to get the answer?
Write and balance the equation.
mols NaOH = M x L = ?
mols HCl = mols NaOH (look at the coefficients in the equation)
M HCl = mols HCl/L HCl
To find the molarity of the hydrochloric acid, you can use the concept of stoichiometry and the balanced chemical equation of the reaction between NaOH and HCl.
The balanced equation is:
NaOH + HCl -> NaCl + H2O
From the balanced equation, you can see that the ratio between NaOH and HCl is 1:1.
Given the volume of NaOH used is 11.4 mL and its molarity is 0.7 M, you can calculate the number of moles of NaOH used:
moles of NaOH = volume (in liters) x molarity = 11.4 mL / 1000 mL/L x 0.7 mol/L = 0.00798 mol
Since the ratio of NaOH to HCl is 1:1, the number of moles of HCl is also 0.00798 mol.
Now, we need to calculate the molarity of HCl. Given the volume of HCl is 5.0 mL, we can convert it to liters:
volume of HCl = 5.0 mL / 1000 mL/L = 0.005 L
Finally, divide the moles of HCl by the volume in liters to calculate the molarity:
molarity of HCl = moles of HCl / volume of HCl = 0.00798 mol / 0.005 L = 1.596 mol/L
So, the molarity of the hydrochloric acid is approximately 1.6 M.
Therefore, the correct answer is b) 1.6.
To find the molarity of the hydrochloric acid, we need to use the concept of stoichiometry in a titration. Here's the step-by-step process to solve this question:
Step 1: Write the balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):
HCl + NaOH → NaCl + H2O
Step 2: Determine the mole ratio between NaOH and HCl from the balanced equation. In this case, it is 1:1, meaning 1 mole of NaOH reacts with 1 mole of HCl.
Step 3: Calculate the number of moles of NaOH used. To do this, we use the formula:
moles = Molarity × volume (in liters)
moles of NaOH = 0.7 M × 11.4 mL ÷ 1000 mL/L
Note: We converted the milliliters (mL) to liters (L) by dividing by 1000.
Step 4: Since the mole ratio is 1:1, the number of moles of NaOH used is equal to the number of moles of HCl present in the 5.0 mL of hydrochloric acid.
Step 5: Calculate the molarity of the hydrochloric acid. To do this, divide the number of moles of HCl by the volume of the acid in liters:
Molarity of HCl = moles of HCl ÷ volume of HCl (in liters)
Molarity of HCl = moles of NaOH ÷ 5.0 mL ÷ 1000 mL/L
Now, let's calculate the molarity of the hydrochloric acid:
Moles of NaOH = 0.7 M × 11.4 mL ÷ 1000 mL/L
Moles of NaOH = 0.00798 mol
Molarity of HCl = 0.00798 mol ÷ 5.0 mL ÷ 1000 mL/L
Molarity of HCl = 1.596 M (rounded to three decimal places)
Therefore, the molarity of the hydrochloric acid is approximately 1.596 M. Comparing this with the options provided, the correct answer is option b) 1.6.