Question

it requires 40 ml of 1m hxa to react completely with 20 ml of 1 m ca(oh)2 what is the value of x

Answer 1

Answer:

Since the reactants react in a 1:1 mole ratio, 0.0020 mol of was used up. Now, n = c*v 0.0020 = 0.1 * v v = 0.02 l v =20 ml

Answer 2

Given: 40ml  of 1M $H_{x} A$ reacts with 20ml of 1M $Ca(OH)_{2}$.

To find: The value of x.

Formula to be used: $N_{1} V_{1}=N_{2} V_{2}$

N=normality and V=volume.

Step-by-step explanation:

Step 1 of 2

When an acid and base undergoes neutralization, $N_{1} V_{1}=N_{2} V_{2}$  can be used.

Here N is normality and V is the volume.

The normality can be calculated by multiplying the Molarity with n-factor.

So, the equation can be written as:

$M_{1} V_{1}n_{1}=M_{2} V_{2}n_{2}$

$H_{x} A$ ⇄ $xH^{+} +A^{x-}$

n-factor for $H_{x} A$= x

n-factor for $Ca(OH)_{2}$ = 2

Step 2 of 2

now substituting the values, we will get:

$M_{1} V_{1}n_{1}=M_{2} V_{2}n_{2}\\\\40*1*x=20*1*2\\40x=40\\x=1$

THEREFORE, VALUE OF X IS 1, THE FORMULA IS HA.