How many moles of NaOH are needed to neutralize 0.010 L of 0.20 M of chromic acid
To determine how many moles of NaOH are needed to neutralize 0.010 L of 0.20 M chromic acid (H2CrO4), we need to first write out the balanced chemical equation for the reaction between chromic acid and sodium hydroxide:
H2CrO4 + 2NaOH -> Na2CrO4 + 2H2O
From the balanced equation, we can see that the mole ratio between chromic acid (H2CrO4) and sodium hydroxide (NaOH) is 1:2.
Given that the concentration of chromic acid is 0.20 M and the volume is 0.010 L, we can calculate the number of moles of chromic acid:
moles of chromic acid = concentration x volume
moles of chromic acid = 0.20 mol/L x 0.010 L
moles of chromic acid = 0.002 moles
Since the mole ratio between chromic acid and sodium hydroxide is 1:2, we need twice as many moles of NaOH to neutralize the chromic acid. Therefore, the number of moles of NaOH needed is:
moles of NaOH = 2 x moles of chromic acid
moles of NaOH = 2 x 0.002 moles
moles of NaOH = 0.004 moles
Therefore, 0.004 moles of NaOH are needed to neutralize 0.010 L of 0.20 M chromic acid.