Solveeit Logo
Login

Question

Question: What is pH of 50 ml 0.05 (M) NaOH and 50 ml 0.1 (N) \(C{H_3}COOH\)....

What is pH of 50 ml 0.05 (M) NaOH and 50 ml 0.1 (N) CH3COOHC{H_3}COOH.

Explanation

Solution

First one should know that pH of the total solution will be determined using the product of above reactants which is a sodium salt of carboxylic acid. Further, we know that
Number of moles = Molarity of the solution ×\timesVolume of the solution (in L)
Using this, we will find the number of moles and then pH.
We know that Ka(CH3COOH)=1.8×10−5{K_a}(C{H_3}COOH) = 1.8 \times {10^{ - 5}}

Complete step by step answer:
First, let us understand the reaction that is occurring between NaOH and CH3COOHC{H_3}COOH.
CH3COOH+NaOH→CH3COONa+H2OC{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O
From the balanced equation, it is clear that one mole of NaOH reacts with one mole of CH3COOHC{H_3}COOH to give salt and water. So, they react in 1:1 ratio.
Now, let us write the things given to us.
So, we have Molarity of NaOH = 0.05 M
Volume of NaOH = 50 mL = 0.05 L
Normality of CH3COOHC{H_3}COOH= 0.1 N
Volume of CH3COOHC{H_3}COOH= 50 mL = 0.05 L
Total Volume of mixture after combining the CH3COOHC{H_3}COOHand NaOH is-
Total volume of mixture = Volume of CH3COOHC{H_3}COOH+ Volume of NaOH
Total volume of mixture = 50 + 50 = 100 mL = 0.1 L
Now, let us calculate the number of moles.
Number of moles = Molarity of the solution ×\timesVolume of the solution (in L)
Number of moles of NaOH = 0.05 M ×\times 0.05 L
Number of moles of NaOH = 0.0025
Number of moles of CH3COOHC{H_3}COOH= 0.1×\times0.05 L
Number of moles of CH3COOHC{H_3}COOH= 0.005
Thus, 0.005 moles of CH3COOHC{H_3}COOHwill react with NaOh giving 0.005 moles of CH3COONaC{H_3}COONa in total 100 mL solution.
Thus, molarity of CH3COONaC{H_3}COONa solution = Number of molesVolume of solution\dfrac{{Number{\text{ of moles}}}}{{Volume{\text{ of solution}}}}
Thus, molarity of CH3COONaC{H_3}COONa solution = 0.0050.1\dfrac{{0.005}}{{0.1}}
Molarity of CH3COONaC{H_3}COONa solution =0.05 M
We know that Ka(CH3COOH)=1.8×10−5{K_a}(C{H_3}COOH) = 1.8 \times {10^{ - 5}}
We can find Kb=10−141.8×10−5{K_b} = \dfrac{{{{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}

Kb=5.56×10−10{K_b} = 5.56 \times {10^{ - 10}}
From the value of Kb{K_b}, we can calculate [OH−][O{H^ - }]
Kb=[OH−]2[CH3COONa]{K_b} = \dfrac{{{{[O{H^ - }]}_2}}}{{[C{H_3}COONa]}}
Putting the value of Kb{K_b} and CH3COONaC{H_3}COONa; we can find out value of [OH−][O{H^ - }].
5.56×10−10=[OH−]20.055.56 \times {10^{ - 10}} = \dfrac{{{{[O{H^ - }]}_2}}}{{0.05}}
[OH−]2=5.56×10−10×0.05{[O{H^ - }]_2} = 5.56 \times {10^{ - 10}} \times 0.05
[OH−]2=2.78×10−11{[O{H^ - }]_2} = 2.78 \times {10^{ - 11}}
[OH−]=5.27×10−6[O{H^ - }] = 5.27 \times {10^{ - 6}}
Further, we know that pOH = - log [OH−][O{H^ - }]
So, pOH = -log(5.27×10−6)(5.27 \times {10^{ - 6}})
pOH = 5.28
We know, pH = 14-pOH
Thus, pH = 14-5.28
pH=8.72

Thus, the pH of the total solution is 8.72.

Note: During calculating the number of moles, the volume of solution is to be taken in L only. If we take the value in mL then the whole numerical will be wrong. Further, the pH of a system is negative logarithm of hydrogen ion concentration. The total value of pH and pOH comes out to be 14. So, the pH scale ranges from 0 to 14.