600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H 2 SO 4. The pH of the mixture is × 10-2. (Nearest integer) [Given log 2 =0.30 log 3 =0.48 log 5 =0.69 log 7 -0.84 log 11 =1.04]

Question

600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H 2 SO 4. The pH of the mixture is × 10-2. (Nearest integer) [Given log 2 =0.30 log 3 =0.48 log 5 =0.69 log 7 -0.84 log 11 =1.04] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Total milimoles of H + =(600 × 0.01)+(400 × 0.01 × 2) =14 [ H +] =(14/1000)=14 × 10-3 pH =3- log 14 =1.86 =186 × 10-2

Q. $600 m L$ of $0.01 M H Cl$ is mixed with $400 m L$ of $0.01 M H_{2} S O_{4}$ . The $p H$ of the mixture is ______ $\times 1 0^{- 2}$ . (Nearest integer)
[Given $lo g 2 = 0.30$
$lo g 3 = 0.48$
$lo g 5 = 0.69$
$lo g 7 - 0.84$
$lo g 11 = 1.04]$

Total milimoles of $H^{+} = (600 \times 0.01) + (400 \times 0.01 \times 2)$
$= 14$
$[H^{+}] = \frac{14}{1000} = 14 \times 1 0^{- 3}$
$p H = 3 - lo g 14$
$= 1.86$
$= 186 \times 1 0^{- 2}$

Q. 600mL of 0.01MHCl is mixed with 400mL of 0.01MH2​SO4​. The pH of the mixture is ______×10−2. (Nearest integer) [Given log2=0.30 log3=0.48 log5=0.69 log7−0.84 log11=1.04]

Total milimoles of H+=(600×0.01)+(400×0.01×2) =14 [H+]=100014​=14×10−3 pH=3−log14 =1.86 =186×10−2

Q. $600 m L$ of $0.01 M H Cl$ is mixed with $400 m L$ of $0.01 M H_{2} S O_{4}$ . The $p H$ of the mixture is ______ $\times 1 0^{- 2}$ . (Nearest integer)
[Given $lo g 2 = 0.30$
$lo g 3 = 0.48$
$lo g 5 = 0.69$
$lo g 7 - 0.84$
$lo g 11 = 1.04]$

Total milimoles of $H^{+} = (600 \times 0.01) + (400 \times 0.01 \times 2)$
$= 14$
$[H^{+}] = \frac{14}{1000} = 14 \times 1 0^{- 3}$
$p H = 3 - lo g 14$
$= 1.86$
$= 186 \times 1 0^{- 2}$