# Acid-Base neutralization w/ more than 1 base.

Here's the question:
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A base stock solution is prepared by dissolving both 1 gram of sodium hydroxide and 2 grams of potassium hydroxide in enough water to prepare 100 mL of solution.

How many millimeters of this base stock solution are required to neutralize 100 mL of 0.50 M $$H_2SO_4 _(_a_q_)$$ ?
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This is what I have so far.
Using Molarity * Volume = moles I get that there are 0.05 mol of $$H_2SO_4 _(_a_q_)$$

Here's where I get stuck. I've never dealt with a problem where the base is made up of 2 components instead of one (i.e. just lithium hydroxide not a combination of lithium hydroxide and aluminum hydroxide). As such I am unsure where to go with this.

I think that I am going to need 0.1 mol of the base stock solution but that's about it.

Any ideas where I can go from here?

The problem might look complicated on the surface, but it is actually quite easy. What do K(OH) and Na(OH) have in common? What happens when you dissolve them both into water?

They're both strong bases that disassociate in water to form OH$$^-$$ and either K$$^+$$ or Na$$^+$$.

But, you would still have different moles of each so I'm still stuck.

How many moles of OH- in 1 grams of NaOH? In 2 grams of KOH? How many moles would there be altogether?

There would be .025 mol of OH in NaOH and .036 mol of OH in KOH.
Here's what I think the solution should be now. See if this is right.

From before. I have 0.05 mol of $$H_2SO_4$$. Based on the reaction:

$$H_2SO_4 + 2XOH -> X_2SO_4 + 2H_2O$$ where X is either Na or K ( I don't think it matters which one because they are both from group 1 with a charge of 1+).

That means 1 mol of $$H_2SO_4$$ needs to react with 2 mol of XOH.

So, I need 0.1 mol of that.

Total mol of XOH that I have now is 0.025 + 0.036 = 0.061 mol.

Molarity XOH = 0.061 mol / 0.1 L = 0.61 M

Thus, the volume needed should be: 0.1 mol / 0.61 M = 0.164 L = 16.4 mL.

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chem_tr
Gold Member
Hello, find how many moles are there in 1 gram of NaOH and 2 grams of KOH first. Then add these to learn the total mole numbers of OH- ions. Neutralize them with 50 millimoles (or 0.05 moles) of H+ ions. This is just that simple. I don't think volumes as large as 16.4 liters will be needed to neutralize such little amounts of base. Please recalculate your reactions. By the way, your mole calculation is correct, I checked it. You are very close to the solution. Just solve a proportion equation, if there are 0.061 moles of base present in 100 mL of solution, then 0.050 moles of acid will require x mL of that solution.

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I got that the answer should be either: 16.4 mL or using 0.5 mol

then the volume should be (0.05 mol) / (0.61 M) = .082 L = 8.2 mL

Here is the question which answer would it be: 16.4 mL or 8.2 mL?

I guess the larger question is don't I have to take into account that according to the reaction: $$H_2SO_4 + 2XOH -> X_2SO_4 + 2H_2O$$ where X is either Na or K ( I don't think it matters which one because they are both from group 1 with a charge of 1+).

I need 2 moles of XOH for every mol of $$H_2SO_4$$ and thus the answer should be 16.4 mL.

Or does it not matter and I can just use the equation $$M_1V_1 = M_2V_2$$ and thus get 8.2 mL?