Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations 1. 60 mL of (M/10 )HCl+40 mL of (M/10 )NaOH 2. 55 mL of (M/10 )HCl+45 mL of (M/10 )NaOH 3. 75 mL of (M/5 )HCl+25 mL of (M/5 )NaOH 4. 100 mL of (M/10 )HCl+100 mL of (M/10 )NaOH pH of which one of them will be equal to 1?

Question

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations 1. 60 mL of (M/10 )HCl+40 mL of (M/10 )NaOH 2. 55 mL of (M/10 )HCl+45 mL of (M/10 )NaOH 3. 75 mL of (M/5 )HCl+25 mL of (M/5 )NaOH 4. 100 mL of (M/10 )HCl+100 mL of (M/10 )NaOH pH of which one of them will be equal to 1? (A) 4 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Meq of HCl =75× (1/5)× 1=15 Meq of NaOH =25× (1/5)× 1=5 Meq of HCl in resulting solution = 10 Normality of [.H+]. in resulting mixture =(10/100)=(1/10) pH=-log[.H+].=-log[(1/10)]=1.0

$4$

$1$

$2$

$3$

Meq of HCl $= 75 \times \frac{1}{5} \times 1 = 15$

Meq of NaOH $= 25 \times \frac{1}{5} \times 1 = 5$

Meq of HCl in resulting solution = 10

Normality of $[H^{+}]$ in resulting mixture $= \frac{10}{100} = \frac{1}{10}$

$p H = - l o g [H^{+}] = - l o g [\frac{1}{10}] = 1.0$

4

1

2

3

Meq of HCl =75×51​×1=15 Meq of NaOH =25×51​×1=5 Meq of HCl in resulting solution = 10 Normality of [H+] in resulting mixture =10010​=101​ pH=−log[H+]=−log[101​]=1.0

$4$

$1$

$2$

$3$

Meq of HCl $= 75 \times \frac{1}{5} \times 1 = 15$

Meq of NaOH $= 25 \times \frac{1}{5} \times 1 = 5$

Meq of HCl in resulting solution = 10

Normality of $[H^{+}]$ in resulting mixture $= \frac{10}{100} = \frac{1}{10}$

$p H = - l o g [H^{+}] = - l o g [\frac{1}{10}] = 1.0$