Question

two litres of 1M Hcl is mixed with 1 litre of 1M NaOH Solution.Calculate strength of salt formed and PH of resulting solution.
Ans : Strength = 0.33M ,PH=0.48​

Answer 1

Answer:

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NAOH

Answer 2

Answer:

The strength of the salt ( $NaCl$ ) formed is $0.33M$.

The pH of the resulting solution is $0.48$.

Explanation:

Given data,

The molarity of $HCl$, M₁ = $1M$

The volume of $HCl$, V₁ = $2L$

The molarity of $NaOH$, M₂ = $1M$

The volume of $NaOH$, V₂ = $1L$

The strength of the salt formed =?

The pH of the resulting solution =?

The reaction follows:

• $HCl + NaOH \rightarrow NaCl + H_{2}O$

Thus, the salt formed is $NaCl$.

And the molarity of $NaCl$ = $M_{3}$

Also, the volume of $NaCl$ = $V_{1} + V_{2}$

Therefore, from the equation;

• $M_{1} V_{1} + M_{2} V_{2} = M_{3}( V_{1} +V_{2} )$
• $1(2) + 1(1) = M_{3} ( 2+1)$
• $M_{3} = \frac{3}{3}$ = $1M$

This means that $1 mole$ of $NaCl$ in $1 litre$ of the solution.

Now, the strength = $\frac{Number of moles ( NaCl )}{total volume}$ = $\frac{1}{3}$ = $0.33M$

Thus, pH = $-log ( 0.33)$

• pH = $-log( \frac{33}{100})$  
• pH = $- ( log 33 - log 100 )$           ( $log 33 = 1.518$ and $log 100 =2$ )
• pH = $- ( 1.518- 2)$ = $0.48$

Hence, the pH of the resulting solution of strength $0.33M$ is $0.48$.