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In the titration of 20.00 mL of 0.500 M HCl by 0.500 M NaOH, calculate the volume of 0.500 M NaOH required to reach a pH of 2.0.

I have tried to change the 20.00 mL into the amount of substance (0.01 moles), then I turned the 2.0 pH into concentration (0.01 M). I then tried to find the total volume, which came out to be 1 L, which creates 980 mL of NaOH. Plugging it back in though doesn't give 2.0 pH. So, I was wondering if anyone knew how to find the volume, or if there was a formula that I could use to solve it.

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  • $\begingroup$ The hydrogen ion concentration of 0.01 M is a great first step. Adding the NaOH does two things: 1) It adds hydroxide, which lowers the amount of hydrogen ion by reacting with it. 2) It adds water, which lowers the concentration of all solutes. $\endgroup$ – Karsten Theis Apr 14 at 21:29
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$$\ce{NaOH + HCl -> NaCl + H2O}$$

Let $x$ Litres be the amount of $\ce{NaOH}$ needed.

$$-\log[\ce{H+}] = 2$$

$$-\log\left(\frac{(0.5 \times 0.02 - x \times 0.5)}{(0.02 + x)}\right) = 2$$

$$x = \frac{49}{2550}$$

$$x \approx 0.0192 \; \text{(3 sig. fig.)}$$

Thus, $\pu{0.0192 L}$ of $\ce{NaOH}$ is needed.

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