How much of a 0.23 M NaOH solution is needed to neutralize 27 mL of a 0.85 M sulfuric acid solution? The units are mL.
Can you please explain why you did what did
Thank you
well, it takes two moles of NaOH to neutralize on mole of sulfuric acid, so
2*.23(Volumebase)=27ml*.85
solve for volumebase
To solve this problem, we can use the concept of stoichiometry, which relates the number of moles of reactants to each other.
First, let's calculate the number of moles of sulfuric acid (H2SO4) in the given solution.
Moles of H2SO4 = concentration (M) × volume (L)
Since the volume is given in milliliters (mL), we need to convert it to liters by dividing by 1000:
Moles of H2SO4 = (0.85 M) × (27 mL ÷ 1000 mL/L) = 0.02295 moles
Sulfuric acid (H2SO4) is a strong diprotic acid, meaning it can donate two protons (H+) in a reaction. Sodium hydroxide (NaOH) is a strong base that can accept one proton (OH-). The balanced chemical equation for the reaction between H2SO4 and NaOH is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, the number of moles of NaOH needed to neutralize the given amount of H2SO4 is half of the moles of H2SO4.
Moles of NaOH needed = 0.02295 moles ÷ 2 = 0.011475 moles
Now, let's calculate the volume of the NaOH solution needed to obtain the desired moles of NaOH.
Volume (L) = moles ÷ concentration (M)
Since the concentration is given in molarity (M), we can use it directly in our calculation without conversion:
Volume (L) = 0.011475 moles ÷ 0.23 M = 0.049891 L
Since the volume needs to be in mL, we can convert it by multiplying by 1000:
Volume (mL) = 0.049891 L × (1000 mL/L) = 49.891 mL
Therefore, approximately 49.891 mL of the 0.23 M NaOH solution is needed to neutralize 27 mL of the 0.85 M sulfuric acid solution.