How to do this?

Question

What volume of 2.50M NaOH is needed to completely neutralize 80.ml of 1.75m HCL?

2.what is the concentration of 500.ml of NaOH if it is neutralized (completely) by 800.ml of 0.50m HCL??

3. what volume of 0.60M HCL is needed to completely neutralize 125ml of 1.40M NaOH?

how do i do this

Answer 1

Hello, Isabel,

We need a balanced equation to determine the ratio of acid and base required for neutralization.

NaOH + HCl = NaCl + H₂O

The molar ratio of acid to base is 1/1. One mole of HCl will neutralize 1 mole of NaOH.

We can calculate the moles of HCl in the first question. We have 80 ml of 1.75M HCl solution. Molar is defined as moles/liter. We'll convert 80ml to liters and then multiply by 1.75 moles/liter to find moles HCl.

(1.75moles/liter)*(0.08 liters) = 0.14 moles HCl. We need to find out how many ml of the 2.5M solution of NaOH would contain the same moles of NaOH. The molar amounts should be equal, according to the balanced equation. We can set up the calculation as follows:

(2.5moles/liter)*X = 0.14 moles, where X is the liters we need to find. Solving for X gives us 0.056 liters, or 56 ml.

1. The other problems can be solved in the same manner. When the balanced equations says it requires 1 mole of acid to neutralize 1 mole of base, we can use a simpler expression:

M1V1 = M2V2, where M and V are concentration (molarity) and volume of the first and second reagents.

The units need to be the same for both sides. Try this on the first problem to see how it works:

(80ml)*(1.75M) = (2.5M)*V2

V2 = 56 ml HCl

Notice that we didn't even need to change our volume unit to liters. We can leave the input in ml and the output will be ml.

For the second question, the calculation would look like this:

M1V1 = M2V2

X*(500ml) = (0.5M)*(800ml)

X = 0.8M

The third can be done the same way:

M1V1 = M2V2

(0.60M)*(V1) = (125ml)*(1.4M)

V1 = 292 ml

Bob