Question

find the concentration of hcl if 10 ml of 0.5 m ca(oh)2 solution is required to titrate 50 ml of hcl till the neutralization point?​

Answer 1

The concentration of HCl of 10 ml of 0.5 m ca(oh)2 solution is required to titrate 50 ml of HCl till the neutralization point is 0.25 M.

Given:

10 ml of 0.5 m ca(oh)2 solution is required to titrate 50 ml of HCl till the neutralization point.

To Find:

The concentration of HCl of 10 ml of 0.5 m ca(oh)2 solution is required to titrate 50 ml of HCl till the neutralization point.

Solution:

To find the concentration of HCl when 10 ml of 0.5 m ca(oh)2 solution is required to titrate 50 ml of HCl till the neutralization point we will follow the following steps:

As we know,

When a solution is titrated against another solution then at the neutralization point the reactant of two solutions has the same stoichiometric amount.

And,

The relation between their concentration and volume is given by:

$\frac{m1v1}{n1} (for \: HCl)= \frac{m2v2}{n2} \: \: for \: ca(oh)2$

Here, m denotes molarity v denotes the volume and n is the n factor.

n factor is the number of electrons lost or gained so,

n1 = 1

n2 = 2

Because calcium loses 2 electrons and forms calcium dication.

Now,

Putting values in the above formula we get,

$\frac{50 \times m1}{1} = \frac{10 \times 0.5}{2}$

$m1 = \frac{10 \times 0.5}{2 \times 50} = 0.25 \: molar$

Henceforth, the concentration of HCl of 10 ml of 0.5 m ca(oh)2 solution is required to titrate 50 ml of HCl till the neutralization point is 0.25 M.

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