Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

What will be the resultant pHpH when 200 mL200\, mL of, an aqueous solution of HCl(pH=2.0)HCl(p H=2.0) is mixed with 300 mL300\, mL of an aqueous solution of NaOH(pH=12)NaOH (pH=12) ?

A

2.699

B

13.3

C

11.301

D

1.33

Answer

11.301

Explanation

Solution

Given, volume of HClHCl solution =200 mL=200\, mL
pHp H of HClHCl solution =2.0=2.0
∴[H+]\therefore\left[H^{+}\right] in HClHCl
solution =1×10−2=1 \times 10^{-2}
∴\therefore Milliequivalents of HCl=200×1×10−2=2HCl =200 \times 1 \times 10^{-2} =2
Similarly, volume of NaOHNaOH solution =300 mL=300\, mL
pHpH of NaOH=12NaOH =12
pOHpOH of NaOH=14−12=2NaOH =14-12=2
[OH−]\left[O H^{-}\right] in NaOH=14−12=2NaOH =14-12=2
[OH−]\left[ OH ^{-}\right] in NaOHNaOH solution =1×10−2=1 \times 10^{-2}
∴\therefore Milliequivalents of NaOH=300×1×10−2=3NaOH =300 \times 1 \times 10^{-2}=3
Since, NaOHNaOH is in excess.
∴\therefore Remaining milliequivalents =[OH−]=\left[ OH ^{-}\right]
=3−2=1=3-2=1
Remaining concentration of [OH−]\left[ OH ^{-}\right]
=1500=2×10−3=\frac{1}{500}=2 \times 10^{-3}
∴[H+]=10−142×10−3=5×10−12\therefore\left[H^{+}\right]=\frac{10^{-14}}{2 \times 10^{-3}}=5 \times 10^{-12}
pH=−log⁡[H+]p H=-\log \left[H^{+}\right]
=−log⁡(5×10−12)=-\log \left(5 \times 10^{-12}\right)
=11.3010=11.3010