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CBSE Class 10 Answered

Calculate pH of a solution which contains 9.9 ml of 1 M HCl and 100 ml of 0.1 M NaOH. 
Asked by acv27joy | 06 Oct, 2018, 09:21: PM
answered-by-expert Expert Answer

The reaction proceeds as :

NaOH + HCl → NaCl + H2O

As we know,

1 mole of HCl reacts with the 1 mole of NaOH 

lets find the moles of solution,

Mole HCl in 9.9 ml of 1 M solution = fraction numerator 9.9 over denominator 1000 end fraction cross times 1 space equals space 0.0099 space mole space HCl

Mole NaOHin 100 ml of 0.1 M solution = 100 over 1000 cross times 0.1 space equals space 0.01 space mole space NaOH

on mixing the 0.0099 mole of HCl will be neutralised = 0.01 - 0.0099 = 1 space cross times space 10 to the power of negative 4 end exponent space mol space NaOH space dissolved space in space 109.9 space ml space solution

hence,  the molarity of NaOH solution = fraction numerator 1 cross times 10 to the power of negative 4 end exponent over denominator 0.1099 end fraction space equals space 9.099 cross times 10 to the power of negative 4 space end exponent straight M

As we find the molarity lets calculate the pH of solution,

pOH space equals space minus log left parenthesis 9.099 cross times 10 to the power of negative 4 end exponent right parenthesis pOH space equals space 3.04 therefore space pH space equals space 14 space minus space pOH space space space space space Ph space equals space 14.0 minus 3.04 space space space space space space pH space equals space 10.46

Answered by Ramandeep | 12 Oct, 2018, 12:57: PM

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