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siniylev [52]
3 years ago
7

For a titration of 25 ml of naoh solution, 23.30 ml of hcl solution if used. what is the molarity of naoh

Chemistry
1 answer:
Eduardwww [97]3 years ago
3 0

<span>HCl<span>(aq)</span>+NaOH<span>(aq)</span>→NaCl<span>(aq)</span>+<span>H2</span>O<span>(l)</span></span>

As you can see here, one mole of acid neutralizes one mole of base.

We use the concentration equation, which states that,

<span>c=<span>nv</span></span>

<span> <span> <span> n is the number of moles </span> <span> v is the volume of solution </span> </span> </span>

Rearranging for moles, we get,

<span>n=c⋅v</span>

So, we have:

<span><span>n<span>NaOH</span></span>=0.1 M⋅0.05 L</span>

<span>=0.005 mol</span>

Since one mole of acid neutralizes one mole of base, then we must have: <span><span>n<span>HCl</span></span>=<span>n<span>NaOH</span></span></span>.

And so,

<span><span>c<span>HCl</span></span>=<span><span>n<span>HCl</span></span><span>v<span>HCl</span></span></span></span>

<span>=<span><span>0.005 mol</span><span>0.03 L</span></span></span>

<span>≈0.17 <span>M</span></span>

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8 0
3 years ago
Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen
alexgriva [62]

Answer:

V_{base}=24.04mL

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

2M_{base}V_{base}=M_{acid}V_{acid}

Therefore, for is to compute the volume of the used base, we proceed as shown below:

V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}

And we plug in to obtain:

V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL

Best regards!

8 0
3 years ago
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