# A question in double titration

Tags:
1. Jun 11, 2017

### Wrichik Basu

1. The problem statement, all variables and given/known data:

A mixed solution of KOH and sodium carbonate required 15ml of (N/20) HCl solution when titrated with phenolphthalein indicator. But the same amount of solution, when titrated with methyl orange as an indicator, required 25ml of the same HCl. The amount of KOH in the solution is:

1. 0.014g
2. 0.14g
3. 0.028g
4. 1.4g

2. Relevant equations:

3. The attempt at a solution:

Let the number of equivalents of HCl be $x$ and that of sodium carbonate be $y$.

So, using phenolphthalein, $$x+y = \frac {15}{1000} × \frac {1}{20}$$
Using methyl orange,

$$x+2y = \frac {25}{1000} × \frac {1}{20}$$
Solving these two equations, I get a value for $x$ with which I'm not getting any option for the weight of KOH. Where am I going wrong?

2. Jun 11, 2017

### Staff: Mentor

Check your math.

Or, if you are sure about it, show what you got.

3. Jun 11, 2017

### Wrichik Basu

Concept is OK?

4. Jun 11, 2017

### Wrichik Basu

Subtracting first equation from second, $$y=1.25×10^{-3} \; - 0.75×10^{-3}$$ $$Or, \quad y=0.5×10^{-3}$$.
So, $$x=0.25×10^{-3}$$
Therefore, weight of KOH=0.014g.

Very sorry, I mistakenly put the weight of NaOH instead of KOH.

Thank you.

5. Jun 11, 2017

### Staff: Mentor

Happens to everyone.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted