What volume of 0.200 M H2SO4 will be needed to neutralize 50mL of 0.178 N NaOH?
i don't even know how to start
The sulfuric acid is .4N .
.4N(volume)=.178N*50ml
solve for volume
3.56mL?
No, not even close.
You want (50*0.178)/0.4 = ??
22.25 ml?
To solve this problem, you can use the concept of stoichiometry, which relates the moles of one substance to another in a balanced chemical equation. Here's how you can approach this problem step by step:
Step 1: Write a balanced chemical equation for the reaction between H2SO4 and NaOH. The equation is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
Step 2: Calculate the number of moles of NaOH using its molarity and volume.
Molarity of NaOH = 0.178 N
Volume of NaOH = 50 mL = 50/1000 L = 0.05 L
Number of moles of NaOH = Molarity x Volume
= 0.178 mol/L x 0.05 L
= 0.0089 moles
Step 3: Use the balanced equation to determine the stoichiometric ratio between NaOH and H2SO4. From the equation, you can see that 2 moles of NaOH react with 1 mole of H2SO4.
Step 4: Calculate the moles of H2SO4 required using the stoichiometric ratio.
Number of moles of H2SO4 = (Number of moles of NaOH) / 2
= 0.0089 moles / 2
= 0.00445 moles
Step 5: The last step is to calculate the volume of H2SO4 using its molarity and the moles of H2SO4.
Molarity of H2SO4 = 0.200 M
Volume of H2SO4 = (Number of moles of H2SO4) / (Molarity of H2SO4)
= 0.00445 mol / 0.200 mol/L
= 0.02225 L = 22.25 mL
Therefore, 22.25 mL of 0.200 M H2SO4 will be needed to neutralize 50 mL of 0.178 N NaOH. Make sure to check your units and significant figures in each step.