Question

The following data were collected at the endpoint of
a titration performed to find the molarity of an HCl
solution.
Volume of acid (HCl) used = 14.4 mL
Volume of base (NaOH) used = 22.4 mL
Molarity of standard base (NaOH) = 0.20 M
What is the molarity of the acid solution?
(a) 1.6 M
. (b) 0.64 M
(c) 0.31 M
(d) 0.13 M​

Answer 1

Answer:mava=mbvb

Explanation:

Ma×14.4=0.20×22.4

Ma=0.20×22.4÷14.4

Ma=0.31M

Option3 is correct

Thank you I hope you understand well

Answer 2

$\boxed{\green{\sf Answer}}$

$\mathsf{M_{HCl} = 0.31M}$

$\boxed{\green{\sf Explanation}}$

Given:

• $M_{NaOH} = 0.20M$
• $V_{HCl} = 14.4 ml$
• $V_{NaOH}= 22.4ml$

To Find:

molarity of HCl

Solution:

w.k.t,

n-factor of

• HCl is 1 (basicity)
• NaOH is 1 (Acidity)

$\boxed{\red{\boxed{\bold{(No. of \: moles \times nfactor)_{acid }= (No. of \: moles \times n-factor)_{base }}}}}$

$\boxed{\blue{\boxed{\bold{No. of \: moles = \dfrac{Molarity \times Volume(in \: ml)}{1000}}}}}$

$\mathsf{\dfrac{M_{HCl} \times V_{HCl}}{1000} \times n =\dfrac{M_{NaOH} \times V_{NaOH}}{1000} \times n}$

$\mathsf{\dfrac{M_{HCl} \times 14.4ml}{\cancel{1000}} \times 1 = \dfrac{0.20 \times 22.4ml}{\cancel{1000}} \times 1}$

$\mathsf{M_{HCl} \times 14.4ml = 0.20 \times 22.4ml}$

$\mathsf{M_{HCl} = \dfrac{4.48}{14.4}}$

$\mathsf{M_{HCl} = 0.31M}$________option(c)

Note:

You can also prefer the formulae

$\mathsf{M_A V_A=M_B V_B }$

$\texttt{\blue{Aravind}\: \red{Reddy}....!}$