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A 1.245 g sample of limestone ($\ce{CaCO3}$) is pulverized and then treated with 30 mL of 0.035 M $\ce{HCl}$. The excess acid requires 11.56 mL of 1.010 M $\ce{NaOH}$ to be neutralized. Calculate the %wt of limestone in the rock.

The first balanced equation would be:

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

Then:

$$\ce{HCl + NaOH -> H2O + NaCl}$$

I figured out the moles of $\ce{CaCO3}$ if that's useful for this problem 1.245 g of $\ce{CaCO3}$ / 100.089 g (molar mass)= 0.0124 M

I have this but I don't know the next steps on how to solve the problem.

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  • $\begingroup$ Please note that the homework policy on this site requires you to show some personal effort. $\endgroup$
    – Klaus-Dieter Warzecha
    Mar 4, 2015 at 20:43
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    $\begingroup$ I have the two balanced equations CaCO3+ 2HCl --> CaCl2+H2O+ CO2 $\endgroup$
    – Mariana
    Mar 4, 2015 at 20:43
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    $\begingroup$ Great. So post them. $\endgroup$
    – Lighthart
    Mar 4, 2015 at 20:45
  • $\begingroup$ Please check the original question. Are you sure about the concentration of the hydrochloric acid? $\endgroup$
    – Klaus-Dieter Warzecha
    Mar 5, 2015 at 5:16

2 Answers 2

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The question is a bit misleading ;)

Your calculation of the moles of $\ce{CaCO3}$ would be correct if the rock sample would be pure 100% calcium carbonate. But then, all the titration would be a waste of time: There's 100 wt-% $\ce{CaCO3}$ in pure $\ce{CaCO3}$.

So, there's apparently something else in the rock sample (that doesn't disturb the titration).

However, what you know for sure is:

  • the stoichiometry of the reaction between $\ce{CaCO3}$ and $\ce{HCl}$. With other words: How many moles of $\ce{HCl}$ are needed to react one mole of $\ce{CaCO3}$

  • concentration and volume of the $\ce{HCl}$ used in excess to neutralize all the $\ce{CaCO3}$

  • concentration and volume of the $\ce{NaOH}$ to neutralize all the $\ce{HCl}$ that is left after reacting all $\ce{CaCO3}$

HINT:

Did you consider to solve the back titration experiment by starting with the second neutralisation step?

If 11.56mL of a 1.010 M $\ce{NaOH}$ was used to neutralize the remaining $\ce{HCl}$, how much was left of it? Compare that with the initial amount...

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  • $\begingroup$ I don't know how to do back titration is there a formula? $\endgroup$
    – Mariana
    Mar 4, 2015 at 21:53
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    $\begingroup$ I think the exercise is making the assumption that there is acid inert components to the rock. $\endgroup$
    – Lighthart
    Mar 4, 2015 at 23:54
  • $\begingroup$ @Lighthart Yes, that's exactly what I tried to explain. $\endgroup$
    – Klaus-Dieter Warzecha
    Mar 5, 2015 at 4:55
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  1. Calculate the amount of excess acid:

$$\mathrm{x mol HCl = 11.56mL NaOH \times \frac{1.010 mol NaOH}{1000mL}\times \frac{1 mol HCl}{1mol NaOH} = 0.01167 mol HCl }$$

  1. Determine the amount of acid originally added:

$$\mathrm{x mol HCl = 30 mL HCl \times \frac{0.035 mol NaOH}{1000mL}= 0.0011 mol HCl }$$

  1. Realize more acid was neutralized than was originally added. There must be an error in the question some where.
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