What is the pH of a mixture obtained by mixing 75 ml M / 5 HCl and 25 ml M / 5 NaOH aqueous solution?

Question

What is the pH of a mixture obtained by mixing 75 ml M / 5 HCl and 25 ml M / 5 NaOH aqueous solution? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Number of meq of HCl =75 × (1/5)=15 Number of meq of NaOH =25 × (1/5)=5 Number of meq of HCl left =15-5 =10 in 100 ml soln. [ H +]=( text Numbers of meq HCI / text Volume of solution ( ml )) =(10/100)=10-1 pH =1 .

Q. What is the $p H$ of a mixture obtained by mixing $75 m l M /5 H Cl$ and $25 m l M /5 N a O H$ aqueous solution?

Number of meq of $H Cl = 75 \times \frac{1}{5} = 15$
Number of meq of $N a O H = 25 \times \frac{1}{5} = 5$
Number of meq of HCl left $= 15 - 5$
$= 10$ in $100 m l$ soln.
$[H^{+}] = \frac{Numbers of meq H C I}{Volume of solution ( m l )}$
$= \frac{10}{100} = 1 0^{- 1}$
$p H = 1.$

Q. What is the pH of a mixture obtained by mixing 75mlM/5HCl and 25mlM/5NaOH aqueous solution?

Number of meq of HCl=75×51​=15 Number of meq of NaOH=25×51​=5 Number of meq of HCl left =15−5 =10 in 100ml soln. [H+]= Volume of solution (ml) Numbers of meq HCI​ =10010​=10−1 pH=1.

Q. What is the $p H$ of a mixture obtained by mixing $75 m l M /5 H Cl$ and $25 m l M /5 N a O H$ aqueous solution?

Number of meq of $H Cl = 75 \times \frac{1}{5} = 15$
Number of meq of $N a O H = 25 \times \frac{1}{5} = 5$
Number of meq of HCl left $= 15 - 5$
$= 10$ in $100 m l$ soln.
$[H^{+}] = \frac{Numbers of meq H C I}{Volume of solution ( m l )}$
$= \frac{10}{100} = 1 0^{- 1}$
$p H = 1.$