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SCORPION-xisa [38]

2 years ago

6

Calculate the pH during the titration of 30.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 29.3 mL of the base have be

en added. Ka of formic acid = 1.8 x 10-4.

1 answer:

pashok25 [27]2 years ago

7 0

Answer:

3.336.

Explanation:

<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>

<em />

So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>

<em></em>

<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>

∵ pH = - log[H⁺].

<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>

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Number of moles = $\frac{given mass}{Molar mass}$

Molar mass of $K_2PtCl_4$ = $2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol$

Number of moles of $K_2PtCl_4$ = $\frac{100 g}{415.093 g/mol} = 0.241 mole$ .

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