Chemistry doubt

Respected members of Brilliant !

I have a doubt in inorganic chemistry.

What is the pH of the resulting solution when 40 ml of HCl reacts with 16ml of NaOH ? (pH of HCl is 1 and pH of NaOH is 14 )

Please post the answer and solution.

Thanking you in advance

#Chemistry #InorganicChemistry #PH

Easy Math Editor

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Comments

Conc. of HCl = 0.1M Conc. of NaOH = 1M No. of moles of HCl= 0.004 No. of moles of NaOH= 0.016 Therefore, 0.004 moles HCl and NaOH will neutralise each other and 0.016-0.004=0.012 moles of NaOH will remain. conc. of NaOH=0.012/0.056=0.2M pOH= -log(0.2)=0.7 ph=13.3

Arpita Karkera - 6 years, 2 months ago

@Manish Dash

The given pH of NaOH is 14, hence it's molarity is $1$ . Similarly, the molarity of HCl is $10^{-1}$ . Now, we have $16$ ml NaOH and $40$ ml HCl, which means:

Millimoles of NaOH= $16$

Millimoles of HCl= $40*10^{-1}=4$

Now, HCl is in deficit, hence it acts as limiting reagent. $4$ millimoles of HCl will react with $4$ millimoles of NaOH to form $4$ millimoles of neutral compound $NaCl$ . Note that $12$ millimoles of NaOH is still in solution.

Now, after adding both solutions, the new volume is $16+40=56\text{ml}$ . In this $56$ ml, there is $12$ millimoles of NaOH, hence, molarity of this solution is:

$M=\frac {12 \times 10^{-3}} {56} \times 1000 = \frac {12} {56}= \frac {3} {14}=[OH^-]$

Now,

$pOH= -\log {[OH^-]}=-\log{\frac {3} {14}}=0.669$ (approx. . logarithm is in base $10$ )

Therefore:

$pH=14-pOH=13.33$

Raghav Vaidyanathan - 6 years, 2 months ago

Thank you very much @Raghav Vaidyanathan