Two solutions are mixed as given below: 55 mL ( M /10) HCl +45 mL ( M /10) NaOH What is the pH of the resulting solution?

Question

Two solutions are mixed as given below: 55 mL ( M /10) HCl +45 mL ( M /10) NaOH What is the pH of the resulting solution? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

H +=5.5 × 10-3 mol , OH -=4.5 × 10-3 mol Excess H +=1 × 10-3 mol Total volume =0.055+0.045=0.1 L [ H +]=(1 × 10-3/0.1)=1 × 10-2 M pH =- log 10[ H +]=- log 10(1 × 10-2)=2

Q. Two solutions are mixed as given below:
$55 m L \frac{M}{10} H Cl + 45 m L \frac{M}{10} N a O H$
What is the $p H$ of the resulting solution?

$H^{+} = 5.5 \times 1 0^{- 3} m o l, O H^{-} = 4.5 \times 1 0^{- 3} m o l$
Excess $H^{+} = 1 \times 1 0^{- 3} m o l$
Total volume $= 0.055 + 0.045 = 0.1 L$
$[H^{+}] = \frac{1 \times 1 0 ^{- 3}}{0.1} = 1 \times 1 0^{- 2} M$
$p H = - lo g_{10} [H^{+}] = - lo g_{10} (1 \times 1 0^{- 2}) = 2$

Q. Two solutions are mixed as given below: 55mL10M​HCl+45mL10M​NaOH What is the pH of the resulting solution?

H+=5.5×10−3mol,OH−=4.5×10−3mol Excess H+=1×10−3mol Total volume =0.055+0.045=0.1L [H+]=0.11×10−3​=1×10−2M pH=−log10​[H+]=−log10​(1×10−2)=2

Q. Two solutions are mixed as given below:
$55 m L \frac{M}{10} H Cl + 45 m L \frac{M}{10} N a O H$
What is the $p H$ of the resulting solution?

$H^{+} = 5.5 \times 1 0^{- 3} m o l, O H^{-} = 4.5 \times 1 0^{- 3} m o l$
Excess $H^{+} = 1 \times 1 0^{- 3} m o l$
Total volume $= 0.055 + 0.045 = 0.1 L$
$[H^{+}] = \frac{1 \times 1 0 ^{- 3}}{0.1} = 1 \times 1 0^{- 2} M$
$p H = - lo g_{10} [H^{+}] = - lo g_{10} (1 \times 1 0^{- 2}) = 2$