A and B are two salts each $100 L$ made by dissolving $4 g$ of $NaOH$ and $9.8$ g of $H_{2} {SO}_{4}$ in water, respectively. What is the pH of the solution obtained from mixing $40 L$ A and $10 L$ B?

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Solution

Step1:
For solution A $NaOH$ (Sodium Hydroxide)
Molecular weight of $NaOH = 23 + 16 + 1 = 40$
Number of moles= $\frac{weight of solution}{Molecular weight of solution}$
Applying the above formula
Number of moles $= \frac{4}{40}$ $= 0.1 mole$
Molarity $= \frac{0.1}{100} = 10^{- 4} M$
Step 2:
For solution B $H_{2} {SO}_{4}$ (sulfuric acid)
Molecular weight of $H_{2} {SO}_{4}$ $= 2 \times 1 + 32 + 16 \times 4 = 98$
Number of moles= $\frac{weight of solution}{Molecular weight of solution}$
Applying the above formula
Number of moles $= \frac{9.8}{98} = 0.1 mole$
Molariy= $\frac{0.1}{100} = 10^{- 4} M$
Step 3:
When A and B are mixed $= 40 + 10 = 50$
Milliequivalance $= volume \times molarity \times acidity or basicity$
Basicity of $NaOH$ $= 1$
Acidity of $H_{2} {SO}_{4}$ $= 2$
Applying the above formula of milliequivalence.
Milliequivalence of $NaOH =$ $40 \times 10^{- 4} \times 1 = 0.04$
Milliequivalence of $H_{2} {SO}_{4} =$ $10 \times 10^{- 4} \times 2 = 0.02$
When Sodium Hydroxide with Sulfuric Acid forms ${Na}_{2} {SO}_{4}$ (Sodium Sulphate).
Leftover moles of ${(OH)}^{-}$ Hydroxide ions= $0.04 - 0.02 = 0.02$
Number of moles of hydroxide ion= $\frac{Moles}{Total volume}$
Applying the above formula
Number of moles of Hydroxide ions $= \frac{0.02}{50}$ $= 4 \times 10^{- 4} M$
$pOH = - \log {(OH)}^{-}$
Apply the above formula
$\begin{array}{rcl} pOH & = & - \log 4 - \log 10^{- 4} \\ = & - 2 \log 2 + 4 \\ = & 2 \times 0.301 + 4 \end{array}$
$\begin{array}{rcl} pOH & = & - \log 4 - \log 10^{- 4} \\ = & - 2 \log 2 + 4 \\ = & 2 \times 0.301 + 4 \end{array}$
$pOH = 3.4$
$pH + pOH = 14$
Putting the value of $pOH$ in the above formula.
$pH = 14 - 3.4 = 10.6$
Therefore, the pH of the solution= $10.6$ .

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A and B are two salts each 100L made by dissolving 4gof NaOH and 9.8g of H2SO4 in water, respectively. What is the pH of the solution obtained from mixing 40LA and 10L B?

A and B are two salts each $100 L$ made by dissolving $4 g$ of $NaOH$ and $9.8$ g of $H_{2} {SO}_{4}$ in water, respectively. What is the pH of the solution obtained from mixing $40 L$ A and $10 L$ B?