Dilute an Acid

I have a liter of distilled water PH7 and I have 31.45% Muriatic Acid, Question how much muriatic acid do I add to achieve a 5% acid solution. Metric or imperial , metric is prefered.

I think this is how to solve it...

n = number of liters of acid solution contains 0.3145n pure acid and 0.6855n water

Amount of pure acid = 0.3145*n added

Amount of water added = 0.6855*n added

Pure acid / pure water = (0.3145*n)/(.6855*n +1) = .05

0.3145*n = .05*(0.6855*n + 1)

(0.3145 - 0.05 * 0.6855)*n = 0.05

n=0.05 / (0.3145 - 0.05 * 0.6855) = 0.178428 liters

31.45% / 5% = 6.29 parts water to 1 part solution...

314.5ml acid * 20 = 6,290ml water - 685.5 water in solution = 5604.5ml additional water to solution

skip to 9:48 for sample problems if you like...

https://www.technologynetworks.com/neuroscience/tools/molaritycalculator

There is a couple of ways to approach the problem. The first is to add to the 1000 ml of water and the second is to pour off some of the 1000 ml and add the 31.45% solution.

First to add solution to 1000 ml

1000 = ml H₂O

x = ml HCL solution

.3145 = concentration of X sol

.05 = concentration of new sol

.3145X/(1000+X)=.05

X=189 ml

Checking:189*.3145=59.44 g HCL

59.44 ml HCL /1189 ml final solution * 100 = 5%

Second to remove solution:

for a .05 solution of 1000 ml = 50g HCL

.3145X=50

X=159 ml of 31.45% solution

Therefore pour out 159 ml of the 1000 ml water and add 159 ml of 31.45% solution.

It depends on whether the <...5%...> is by volume, by weight, or by molar quantity, which has been withheld from the forum.

Thank You PW, that was what I was looking for. Makes a difference.

Yes, but chemicals are supplied as % w/w, as I said in #11. Eg caustic soda 46% w/w NaOH, sulphuric acid 96% w/w H2SO4. Perhaps the OP will come back and comment.

In my world we have several ways to "measure" depending on what you are trying to achieve in the final result. God knows I wish it were simple percentage.

Does it make a difference? He said the solution he was using was 31.45% HCL.

The problem then becomes a simple dilution/concentration calculation. The one posing the question knows whether it's by weight/volume/molar concentration. So calculating as a straight forward volume to volume would seem to be perfectly valid because in the end we should arrive at the same result. You can reframe the problem as "How much water do we add to a volume of 31.45% HCL to reduce it to 5%." The result will be in whatever units the original solution was made from.

That's right, but almost certainly 31.45% HCl is % w/w

I stand corrected. After reflecting on my previous statement I returned to my original approach and used 34.15 g-HCL/100 g-sol and recalculated using the 189 g of the 34.15% solution and discovered, much to my chagrin, I had a 5.4% solution by weight and not a 5%. Thanks for pointing it out; so I guess I'll just limp on out and retire from the discussion...

No problem! Glad we've sorted it

1 mole muriatic acid solid = 24.19 ml = 36.46 grams = Molar volume: 24.194 cm³/mol

https://www.aqua-calc.com/calculate/mole-to-volume-and-weight/substance/muriatic-blank-acid-coma-and-blank-solid

<...achieve...>

To one significant digit, like the <...5%...>, or to four, like the <...31.45%...>?

100cc of 31.45% acid contains 31.45cc of concentrated acid and 68.55 cc of water.

To dilute to 5% it needs 19x the amount of water to acid ie. 597.55 cc to 31.45cc acid.

So you have to add 597.55-68.55 = 529cc to the original 100cc to make 560.45cc diluted to 5%

It's usually called hydrochloric acid, HCl nowadays, but each to his own!

Assuming the 31.45% and 5% are % w/w, as is usual in the chemical industry.

1 litre water has mass 1000 gm. If x gm as HCl are added to 1 litre, x/(1000+x/31.45%) = 5%. That gives x = 59.45 gm as HCl, 59.45/31.45% = 189 gm acid as supplied. Not 189 ml, as Rich said in #6.

As you're more likely to measure it by volume, density of 31.45% acid = 1.16 gm/ml, so needs 189/1.16 = 163 ml.

Out of interest, what is this <...5% acid solution...> going to be used for, please?

It will be used to activate another solution for an end product. Details not permitted at this time.

~ 4 shots...

I'll drink to that!

Thanks every one and all. I got my answer and it was probably closer than what I was doing. It appears to me any way a tad more involved than I realized.

It was implicit in some answers but to state it explicitly: dilute acid into the water rather than water into the acid.

I used to call on the chemical compounding industry and plenty of workers have been hurt by not remembering this. Another common error is adding water to concentrated caustic.

Are you trying to obtain a certain Ph level?

If so, the best way might be to titrate the mixture using a Ph meter or precision Ph test paper.

Add 1 ml. each step, and stir and measure after each addition once you get close to the target value.

Your distilled water may not respond linearly.

This is why I love my Galaxy Note

So add 189 grams of the acid.