Calculate the pH of mixture of 800 mL of 1-100 M Ca OH 2- 800 mL of 1-40 M HNO 3 and 400 mL of water-A- 2-7B- 3C- 3-2D- 3-7

The correct option is A 2.7[H^+]= ( 800 ×1/40 ) ( 800 ×1/100× 2 )/2 × 10^3 =4/2× 10^ 3=2 × 10^ 3 pH = log [H^+]= log (2 × 10^ 3) = 3 – log2 = 3 – 0.3 = 2.7 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Properties of Water

Calculate the...

Question

Calculate the pH of mixture of 800 mL of $\frac{1}{100} M C a (O H)_{2}$ , 800 mL of $\frac{1}{40} M H N O_{3}$ and 400 mL of water.

3.7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.7

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 2.7
$[H^{+}] = \frac{(800 \times \frac{1}{40}) - (800 \times \frac{1}{100} \times 2)}{2 \times 10^{3}} = \frac{4}{2} \times 10^{- 3} = 2 \times 10^{- 3} p H = - l o g [H^{+}] = - l o g (2 \times 10^{- 3})$
= 3 – log2 = 3 – 0.3 = 2.7

Suggest Corrections

Similar questions

Q. Calculate pH mixture of

(400 m l, \frac{1}{200} M H_{2} S O_{4}) + (400 m l, \frac{1}{100} M HCl) + (200 m l of water)

Q. Calculate pH of mixture of

(400 m L, \frac{1}{200} M B a (O H)_{2}) + (400 m L, \frac{1}{50} M H C l) + (200 mL of water)

l o g 2 = 0.3010

Q. Calculate pH of mixture of

(400 m L, \frac{1}{200} M B a (O H)_{2}) + (400 m L, \frac{1}{50} M H C l) + (200 mL of water)

l o g 2 = 0.3010

Q. Calculate the pH of a solution obtained by mixing

100 m L

of an acid of

p H = 3

and

400 m L

solution of

p H = 1

Q. Equal volumes of three acid solutions of

p H 3, 4, a n d 5

are mixed in a vessel. What will be the

H^{+}

ion concentration of the mixture ?

a n t i l o g_{b} (- x) = b^{- x}

Join BYJU'S Learning Program

Calculate the pH of mixture of 800 mL of 1100 M Ca(OH)2, 800 mL of 140 M HNO3 and 400 mL of water.

The correct option is B 2.7[H+]=(800×140)−(800×1100×2)2×103=42×10−3=2×10−3pH=−log[H+]=−log(2×10−3) = 3 – log2 = 3 – 0.3 = 2.7

Calculate the pH of mixture of 800 mL of $\frac{1}{100} M C a (O H)_{2}$ , 800 mL of $\frac{1}{40} M H N O_{3}$ and 400 mL of water.

The correct option is B 2.7
$[H^{+}] = \frac{(800 \times \frac{1}{40}) - (800 \times \frac{1}{100} \times 2)}{2 \times 10^{3}} = \frac{4}{2} \times 10^{- 3} = 2 \times 10^{- 3} p H = - l o g [H^{+}] = - l o g (2 \times 10^{- 3})$
= 3 – log2 = 3 – 0.3 = 2.7