How many mL of a 3.2M HCl solution would be needed to neutralize 150mL of 2.6M NaOH? Both NaOH and HCl are ‘strong’ acids/bases.

These react in 1:1 ratio; therefore,

mLacid x Macid = mLbase x Mbase

To determine the amount of HCl solution needed to neutralize the NaOH solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and NaOH.

The balanced chemical equation for the reaction is:

HCl + NaOH -> NaCl + H2O

From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.

First, let's determine the number of moles of NaOH in the 150 mL of 2.6 M NaOH solution:

Moles of NaOH = Volume (L) x Concentration (M)
= 150 mL x (1 L / 1000 mL) x 2.6 M
= 0.39 moles

According to the stoichiometry, the number of moles of HCl required to neutralize the NaOH is equal to the number of moles of NaOH.

Therefore, we need 0.39 moles of HCl.

Now, let's calculate the volume (in mL) of 3.2 M HCl solution that contains 0.39 moles of HCl:

Volume (mL) = Moles of HCl / Concentration (M)
= 0.39 moles / 3.2 M
≈ 122 mL

Therefore, approximately 122 mL of the 3.2 M HCl solution would be needed to neutralize 150 mL of the 2.6 M NaOH solution.